NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

 

NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions: We can see many patterns in nature such as the arrangement of petals of a flower, the pattern of honeycomb, etc. Solutions of NCERT class 10 maths chapter 5 Arithmetic Progressions will give you assistance while doing homework as well as while preparing for the examinations by using such practical examples. We follow certain patterns in daily life. For example, Seema puts rupees 1000 into her daughter’s money box when she was one year old and increased the amount by 100 every year. Then the pattern will be 1000,1100,1200,...........for 1st, 2nd, 3rd,................years. In CBSE NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions, we will explain such type of questions in which succeeding terms are obtained by adding the same number to the preceding terms. In NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions we will also study how to find their general term and the sum of n consecutive terms in the pattern mentioned above, and usage of this knowledge in solving some daily life problems. You can download the NCERT solutions by clicking on the link.

The sum of first n terms is given by

S_n=\frac{n}{2}(2a+(n-1)d)=\frac{n}{2}(first\ term+last\ term)

where a is the first term, n is the number of terms in the given A.P and d is a common difference.

Types of questions asked from class 10 maths chapter 5 Arithmetic Progression

  • Checking whether the series is an arithmetic progression or not.

  • To find the nth term or last term or an arithmetic progression.

  • Questions  based on the sum of an arithmetic progression

  • Use of linear equation concept in an arithmetic progression

  • Application of nth term and sum formulae

 

NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 5.1

Q1 (i) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is  \small Rs\hspace{1mm}15  for the first km and \small Rs\hspace{1mm}8  for each additional  \small km.

Answer:

It is given that
Fare for  1^{st} \ km = Rs. 15
And after that \small Rs\hspace{1mm}8  for each additional  \small km
Now,
Fare for  2^{nd} \ km = Fare of first km + Additional fare for 1 km 
                               =  Rs. 15 + 8 = Rs 23

Fare for 3^{rd} \ km = Fare of first km + Fare of additional second km + Fare of additional third km 
                              = Rs. 23 + 8= Rs 31

Fare of n km =  15 + 8 \times (n - 1) 
( We multiplied by n - 1 because the first km was fixed and for rest, we are adding additional fare.

In this, each subsequent term is obtained by adding a fixed number (8) to the previous term.)

Now, we can clearly see that this is an A.P. with the first term (a) = 15 and  common difference (d) = 8

 

Q1 (ii) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (ii) The amount of air present in a cylinder when a vacuum pump removes   \small \frac{1}{4}  of the air  remaining in the cylinder at a time.

Answer:

It is given that 
vacum pump removes   \small \frac{1}{4}  of the air  remaining in the cylinder at a time
Let us take initial quantity of air = 1

Now, the quantity of air removed in first step = 1/4

Remaining quantity after 1st step

= 1-\frac{1}{4}= \frac{3}{4}

Similarly, Quantity removed after 2nd step = Quantity removed in first step \times  Remaining quantity after 1st step

=\frac{3}{4}\times \frac{1}{4}= \frac{3}{16}
Now,

Remaining quantity after 2nd step would be = Remaining quantity after 1st step - Quantity removed after 2nd step

=\frac{3}{4}- \frac{3}{16}= \frac{12-3}{16}= \frac{9}{16}
Now, we can clearly see that

After the second step the difference between second and first and first and initial step is not the same, hence

the common difference (d)  is not the same after every step 

Therefore, it is not an AP

Q1 (iii) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (iii) The cost of digging a well after every meter of digging, when it costs  \small Rs\hspace{1mm}150  for the first metre and rises by  \small Rs\hspace{1mm}50  for each subsequent meter.

Answer:

It is given that
Cost of digging of 1st meter = Rs 150
and
rises by  \small Rs\hspace{1mm}50  for each subsequent meter
Therefore,

Cost of digging of first 2 meters = cost of digging of first meter + cost of digging additional meter

Cost of digging of first 2 meters = 150 + 50

                                                   =  Rs 200

Similarly,
Cost of digging of first 3 meters = cost of digging of first 2 meters + cost of digging of additional meter

Cost of digging of first 3 meters = 200 + 50

                                                  =  Rs 250

We can clearly see that 150, 200,250, ... is in AP with each subsequent term is obtained by adding a fixed number (50) to the previous term.

Therefore, it is an AP with first term (a) = 150 and common difference (d) = 50

Q1 (iv) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (iv) The amount of money in the account every year, when  \small Rs\hspace{1mm}10000  is deposited at  compound interest at  \small 8\hspace{1mm}\%  per annum.

Answer:

Amount in the beginning = Rs. 10000

Interest at the end of 1st year at the rate of \small 8\hspace{1mm}\%
is \small 8\hspace{1mm}\% of 10000 = \frac{8\times 10000}{100}= 800

Therefore, amount at the end of 1st year will be

= 10000 + 800

= 10800

Now,

Interest at the end of 2nd year at rate of \small 8\hspace{1mm}\%
is  \small 8\hspace{1mm}\% of 10800 =\frac{8\times 10800}{100}= 864

Therefore,, amount at the end of 2ndyear

= 10800 + 864 = 11664

Since each subsequent term is not obtained by adding a unique number to the previous term; hence, it is not an AP

Q2 (i) Write first four terms of the AP, when the first term a and the common difference d are given as  follows \small a=10,d=10

Answer:

It is given that 
\small a=10,d=10
Now,
a_1= a =10
a_2= a_1 + d= 10 + 10 = 20
a_3= a_2 + d= 20 + 10 = 30
a_4= a_3 + d= 30 + 10 = 40
Therefore, the first four terms of the given series are 10,20,30,40 

Q2 (ii) Write first four terms of the AP when the first term a and the common difference d are given as follows:\small a=-2,d=0
Answer:

It is given that 
\small a=-2,d=0
Now,
a_1= a = -2
a_2= a_1 + d= -2 + 0 = -2
a_3= a_2 + d= -2 + 0 = -2
a_4= a_3 + d= -2 + 0 = -2
Therefore, the first four terms of the given series are -2,-2,-2,-2

Q2 (iii) Write first four terms of the AP when the first term a and the common difference d are given as follows \small a=4,d=-3

Answer:

It is given that 
\small a=4,d=-3
Now,
a_1= a =4
a_2= a_1 + d= 4 - 3 = 1
a_3= a_2 + d= 1 - 3 = -2
a_4= a_3 + d= -2- 3 = -5
Therefore, the first four terms of the given series are 4,1,-2,-5

Q2 (iv) Write first four terms of the AP when the first term a and the common difference d are given  as follows \small a=-1,d=\frac{1}{2}

Answer:

It is given that 
\small a=-1,d=\frac{1}{2}
Now,
a_1= a =-1
a_2= a_1 + d= -1 + \frac{1}{2} = -\frac{1}{2}
a_3= a_2 + d= -\frac{1}{2} + \frac{1}{2} = 0
a_4= a_3 + d= 0+\frac{1}{2}= \frac{1}{2}
Therefore, the first four terms of the given series are  -1,-\frac{1}{2},0, \frac{1}{2}

Q2 (v) Write first four terms of the AP when the first term a and the common difference d are given as follows \small a=-1.25,d=-0.25

Answer:

It is given that 
\small a=-1.25,d=-0.25
Now,
a_1= a =-1.25
a_2= a_1 + d= -1.25 -0.25= -1.50
a_3= a_2 + d= -1.50-0.25=-1.75
a_4= a_3 + d= -1.75-0.25=-2
Therefore, the first four terms of the given series are  -1.25,-1.50,-1.75,-2

Q3 (i) For the following APs, write the first term and the common difference:  \small 3,1,-1,-3,...

Answer:

Given AP series is

\small 3,1,-1,-3,...

Now, first term of this AP series is 3

Therefore,

First-term of AP series (a) = 3

Now,

a_1=3 \ \ and \ \ a_2 = 1

And common difference (d)  = a_2-a_1 = 1-3 = -2

Therefore, first term and common difference is 3 and -2 respectively 

Q3 (ii) For the following APs, write the first term and the common difference: \small -5,-1,3,7,...

Answer:

Given AP series is

\small -5,-1,3,7,...

Now, the first term of this AP series is -5

Therefore,

First-term of AP series (a) = -5

Now,

a_1=-5 \ \ and \ \ a_2 = -1

And common difference (d)  = a_2-a_1 = -1-(-5) = 4

Therefore, the first term and the common difference is -5 and  4 respectively 

Q3 (iii) For the following APs, write the first term and the common difference: \small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...

Answer:

Given AP series is

\small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...
Now, the first term of this AP series is \frac{1}{3}

Therefore,

The first term of AP series (a) = \frac{1}{3}

Now,

a_1=\frac{1}{3} \ \ and \ \ a_2 = \frac{5}{3}

And common difference (d)  = a_2-a_1 = \frac{5}{3}-\frac{1}{3} = \frac{5-1}{3} =\frac{4}{3}

Therefore, the first term and the common difference is \frac{1}{3} and  \frac{4}{3} respectively 

Q3 (iv) For the following APs, write the first term and the common difference: \small 0.6,1.7,2.8,3.9,...

Answer:

Given AP series is

\small 0.6,1.7,2.8,3.9,...

Now, the first term of this AP series is 0.6

Therefore,

First-term of AP series (a) = 0.6

Now,

a_1=0.6 \ \ and \ \ a_2 = 1.7

And common difference (d)  = a_2-a_1 = 1.7-0.6 = 1.1

Therefore, the first term and the common difference is 0.6 and  1.1 respectively.

Q4 (i) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small 2,4,8,12,...

Answer:

Given series is
\small 2,4,8,12,...
Now,
the first term to this series is = 2
Now,
a_1 = 2 \ \ and \ \ a_2 = 4 \ \ and \ \ a_3 = 8
a_2-a_1 = 4-2 = 2
a_3-a_2 = 8-4 = 4
We can clearly see that the difference between terms are not equal 
Hence, given series is not an AP

Q4 (ii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small 2,\frac{5}{2},3,\frac{7}{2},...

Answer:

Given series is
\small 2,\frac{5}{2},3,\frac{7}{2},...
Now,
first term to this series is = 2
Now,
a_1 = 2 \ \ and \ \ a_2 = \frac{5}{2} \ \ and \ \ a_3 = 3 \ \ and \ \ a_4 = \frac{7}{2}
a_2-a_1 = \frac{5}{2}-2 = \frac{5-4}{2}=\frac{1}{2}
a_3-a_2 = 3-\frac{5}{2} = \frac{6-5}{2} = \frac{1}{2}
a_4-a_3=\frac{7}{2}-3=\frac{7-6}{2} =\frac{1}{2}
We can clearly see that the difference between terms are equal  and equal to \frac{1}{2}
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = \frac{7}{2}+\frac{1}{2} = \frac{8}{2}=4
a_6=a_5+d = 4+\frac{1}{2} = \frac{8+1}{2}=\frac{9}{2}
a_7=a_6+d =\frac{9}{2} +\frac{1}{2} = \frac{10}{2}=5

Therefore, next three terms of given series are  4,\frac{9}{2} ,5

Q4 (iii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small -1.2,-3.2,-5.2,-7.2,...

Answer:

Given series is
\small -1.2,-3.2,-5.2,-7.2,...
Now,
the first term to this series is = -1.2
Now,
a_1 = -1.2 \ \ and \ \ a_2 = -3.2 \ \ and \ \ a_3 = -5.2 \ \ and \ \ a_4 = -7.2
a_2-a_1 = -3.2-(-1.2) =-3.2+1.2=-2
a_3-a_2 = -5.2-(-3.2) =-5.2+3.2 = -2
a_4-a_3=-7.2-(-5.2)=-7.2+5.2=-2
We can clearly see that the difference between terms are equal  and equal to -2
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = -7.2-2 =-9.2
a_6=a_5+d = -9.2-2 =-11.2
a_7=a_6+d = -11.2-2 =-13.2
Therefore, next three terms of given series are  -9.2,-11.2,-13.2

Q4 (iv) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small -10,-6,-2,2,...

Answer:

Given series is
\small -10,-6,-2,2,...
Now,
the first term to this series is = -10
Now,
a_1 = -10 \ \ and \ \ a_2 = -6 \ \ and \ \ a_3 = -2 \ \ and \ \ a_4 = 2
a_2-a_1 = -6-(-10) =-6+10=4
a_3-a_2 = -2-(-6) =-2+6 = 4
a_4-a_3=2-(-2)=2+2=4
We can clearly see that the difference between terms are equal  and equal to 4
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = 2+4 =6
a_6=a_5+d = 6+4=10
a_7=a_6+d = 10+4=14
Therefore, next three terms of given series are  6,10,14

Q4 (v) Which of the following are APs? If they form an AP, find the common difference d and write three more terms.  \small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...

Answer:

Given series is
\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...
Now,
the first term to this series is = 3
Now,
a_1 = 3 \ \ and \ \ a_2 = 3+\sqrt2 \ \ and \ \ a_3 = 3+2\sqrt2 \ \ and \ \ a_4 = 3+3\sqrt2
a_2-a_1 = 3+\sqrt2-3= \sqrt2
a_3-a_2 = 3+2\sqrt2-3-\sqrt2 = \sqrt2
a_4-a_3 = 3+3\sqrt2-3-2\sqrt2 = \sqrt2
We can clearly see that the difference between terms are equal  and equal to \sqrt2
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = 3+3\sqrt2+\sqrt2=3+4\sqrt2
a_6=a_5+d = 3+4\sqrt2+\sqrt2=3+5\sqrt2
a_7=a_6+d = 3+5\sqrt2+\sqrt2=3+6\sqrt2
Therefore, next three terms of given series are  3+4\sqrt2, 3+5\sqrt2,3+6\sqrt2

Q4 (vi) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small 0.2,0.22,0.222,0.2222,...

Answer:

Given series is
\small 0.2,0.22,0.222,0.2222,...
Now,
the first term to this series is = 0.2
Now,
a_1 = 0.2 \ \ and \ \ a_2 = 0.22 \ \ and \ \ a_3 = 0.222 \ \ and \ \ a_4 = 0.2222
a_2-a_1 = 0.22-0.2=0.02
a_3-a_2 = 0.222-0.22=0.002

We can clearly see that the difference between terms are not equal  
Hence, given series is not an AP

Q4 (vii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small 0,-4,-8,-12,...

Answer:

Given series is
\small 0,-4,-8,-12,...
Now,
first term to this series is = 0
Now,
a_1 = 0 \ \ and \ \ a_2 = -4 \ \ and \ \ a_3 = -8 \ \ and \ \ a_4 = -12
a_2-a_1 = -4-0 =-4
a_3-a_2 = -8-(-4) =-8+4 = -4
a_4-a_3=-12-(-8)=-12+8=-4
We can clearly see that the difference between terms are equal  and equal to -4
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = -12-4 =-16
a_6=a_5+d = -16-4=-20
a_7=a_6+d = -20-4=-24
Therefore, the next three terms of given series are  -16,-20,-24

Q4 (viii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. \small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...

Answer:

Given series is
\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...
Now,
the first term to this series is = -\frac{1}{2}
Now,
a_1 = -\frac{1}{2} \ \ and \ \ a_2 = -\frac{1}{2} \ \ and \ \ a_3 = -\frac{1}{2} \ \ and \ \ a_4 = -\frac{1}{2}
a_2-a_1 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
a_3-a_2 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
a_4-a_3 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
We can clearly see that the difference between terms are equal  and equal to 0
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = -\frac{1}{2}+0=-\frac{1}{2}
a_6=a_5+d = -\frac{1}{2}+0=-\frac{1}{2}
a_7=a_6+d = -\frac{1}{2}+0=-\frac{1}{2}
Therefore, the next three terms of given series are  -\frac{1}{2},-\frac{1}{2},-\frac{1}{2}

Q4 (ix) Which of the following are APs? If they form an AP, find the common difference d and write three more terms.  \small 1,3,9,27,...

Answer:

Given series is
\small 1,3,9,27,...
Now,
the first term to this series is = 1
Now,
a_1 = 1 \ \ and \ \ a_2 = 3 \ \ and \ \ a_3 = 9 \ \ and \ \ a_4 = 27
a_2-a_1 = 3-1=2
a_3-a_2 =9-3=6


We can clearly see that the difference between terms are not  equal  
Hence, given series is not an AP

Q4 (x) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small a,2a,3a,4a,...

Answer:

Given series is
\small a,2a,3a,4a,...
Now,
the first term to this series is = a
Now,
a_1 = a \ \ and \ \ a_2 = 2a \ \ and \ \ a_3 = 3a \ \ and \ \ a_4 = 4a
a_2-a_1 = 2a-a =a
a_3-a_2 = 3a-2a =a
a_4-a_3=4a-3a=a
We can clearly see that the difference between terms are equal  and equal to a
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d =4a+a=5a
a_6=a_5+d =5a+a=6a
a_7=a_6+d =6a+a=7a
Therefore, next three terms of given series are  5a,6a,7a

Q4 (xi) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.  \small a,a^2,a^3,a^4,...
Answer:

Given series is
\small a,a^2,a^3,a^4,...
Now,
the first term to this series is = a
Now,
a_1 = a \ \ and \ \ a_2 = a^2 \ \ and \ \ a_3 = a^3 \ \ and \ \ a_4 = a^4
a_2-a_1 = a^2-a =a(a-1)
a_3-a_2 = a^3-a^2 =a^2(a-1)

We can clearly see that the difference between terms are not equal  
Hence, given series is not in AP

Q4 (xii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. \small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...

Answer:

Given series is
\small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...
We can rewrite it as 
\sqrt2,2\sqrt2,3\sqrt2,4\sqrt2,....
Now,
first term to this series is = a
Now,
a_1 = \sqrt2 \ \ and \ \ a_2 = 2\sqrt2 \ \ and \ \ a_3 = 3\sqrt2 \ \ and \ \ a_4 = 4\sqrt2
a_2-a_1 = 2\sqrt2-\sqrt2 =\sqrt2
a_3-a_2 = 3\sqrt2-2\sqrt2 =\sqrt2
a_4-a_3=4\sqrt2-3\sqrt2=\sqrt2
We can clearly see that difference between terms are equal  and equal to \sqrt2
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d =4\sqrt2+\sqrt2=5\sqrt2
a_6=a_5+d =5\sqrt2+\sqrt2=6\sqrt2
a_7=a_6+d =6\sqrt2+\sqrt2=7\sqrt2
Therefore, next three terms of given series are 5\sqrt2,6\sqrt2,7\sqrt2

That is the next three terms are \sqrt{50},\ \sqrt{72},\ \sqrt{98}

Q4 (xiii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms.  \small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...

Answer:

Given series is
\small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...
Now,
the first term to this series is = \sqrt3
Now,
a_1 = \sqrt3 \ \ and \ \ a_2 = \sqrt6 \ \ and \ \ a_3 = \sqrt9 \ \ and \ \ a_4 = \sqrt{12}
a_2-a_1 = \sqrt6-\sqrt3 =\sqrt3(\sqrt2-1)
a_3-a_2 = 3-\sqrt3 =\sqrt3(\sqrt3-1)
We can clearly see that the difference between terms are not equal  
Hence, given series is not in AP

Q4 (xiv) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.  \small 1^2,3^2,5^2,7^2,...

Answer:

Given series is
\small 1^2,3^2,5^2,7^2,...
we can rewrite it as
1,9,25,49,....
Now,
the first term to this series is = 1
Now,
a_1 =1 \ \ and \ \ a_2 = 9 \ \ and \ \ a_3 =25 \ \ and \ \ a_4 = 49
a_2-a_1 = 9-1 = 8
a_3-a_2 = 25-9=16
We can clearly see that the difference between terms are not equal  
Hence, given series is not in AP

Q4 (xv) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.  \small 1^2,5^2,7^2,73,...

Answer:

Given series is
\small 1^2,5^2,7^2,73,...
we can rewrite it as
1,25,49,73....
Now,
the first term to this series is = 1
Now,
a_1 =1 \ \ and \ \ a_2 = 25 \ \ and \ \ a_3 =49 \ \ and \ \ a_4 = 73
a_2-a_1 = 25-1 = 24
a_3-a_2 = 49-25=24
a_4-a_3 = 73-49=24
We can clearly see that the difference between terms are  equal and equal to 24  
Hence, given series is  in AP
Now, the next three terms are
a_5=a_4+d = 73+24=97
a_6=a_5+d = 97+24=121
a_7=a_6+d = 121+24=145
Therefore, the next three terms of given series are 97,121,145

NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 5.2

Q1 Fill in the blanks in the following table, given that a is the first term, d the common  difference and \small a_n  the \small nth term of the AP:                                                                             

 

a

d

n

\small a_n

(i)

(ii)

(iii)

(iv)

(v)

7

\small -18

\small ...

\small -18.9

\small 3.5

3

\small ...

\small -3

\small 2.5

0

8

10

18

\small ...

105

\small ...

0

\small -5

\small 3.6

\small ...

Answer:

(i)
It is given that 
a=7, d = 3 , n = 8
Now,  we know that
a_n = a+(n-1)d
a_8 = 7+(8-1)3= 7+7\times 3 = 7+21 = 28
Therefore, 
a_8 = 28

(ii) It is given that 
a=-18, n = 10, a_{10} = 0
Now,  we know that
a_n = a+(n-1)d
a_{10} = -18+(10-1)d
0 +18=9d
d = \frac{18}{9}=2
(iii) It is given that 
d=-3, n = 18, a_{18} = -5
Now,  we know that
a_n = a+(n-1)d
a_{18} = a+(18-1)(-3)
-5=a+17\times (-3)
a = 51-5 = 46
Therefore, 
a = 46

(iv) It is given that 
a=-18.9, d = 2.5, a_{n} = 3.6
Now,  we know that
a_n = a+(n-1)d
a_{n} = -18.9+(n-1)2.5
3.6+18.9= 2.5n-2.5
n = \frac{22.5+2.5}{2.5}= \frac{25}{2.5}= 10
Therefore, 
n = 10

(v) It is given that 
a=3.5, d = 0, n = 105
Now,  we know that
a_n = a+(n-1)d
a_{105} = 3.5+(105-1)0
a_{105} = 3.5
Therefore, 
a_{105} = 3.5

Q2 (i) Choose the correct choice in the following and justify:  \small 30th term of the AP:  \small 10,7,4,...,   is

(A) \small 97             (B) \small 77             (C) \small -77             (D) \small -87

Answer:

Given series is 
\small 10,7,4,...,
Here, a = 10
and 
d = 7 - 10 = -3
Now, we know that
a_n = a+(n-1)d
It is given that n = 30
Therefore,
a_{30} = 10+(30-1)(-3)
a_{30} = 10+(29)(-3)
a_{30} = 10-87 = -77
Therefore, \small 30th term of the AP:  \small 10,7,4,...,   is -77
Hence, Correct answer is (C)

Q2 (ii) Choose the correct choice in the following and justify : 11th term of the AP:  \small -3,-\frac{1}{2},2,...,  is  

(A) \small 28              (B) \small 22              (C)  \small -38             (D)  \small -48\frac{1}{2}

Answer:

Given series is 
\small -3,-\frac{1}{2},2,...,
Here, a = -3
and 
d =-\frac{1}{2} -(-3)= -\frac{1}{2} + 3 = \frac{-1+6}{2}= \frac{5}{2}
Now, we know that
a_n = a+(n-1)d
It is given that n = 11
Therefore,
a_{11} = -3+(11-1)\left ( \frac{5}{2} \right )
a_{11} = -3+(10)\left ( \frac{5}{2} \right )
a_{11} = -3+5\times 5 = -3+25 = 22
Therefore, 11th term of the AP: \small -3,-\frac{1}{2},2,...,    is 22
Hence, the Correct answer is (B)

 

Q3 (i) In the following APs, find the missing terms in the boxes :  \small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26

Answer:

Given AP series is
\small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26
Here, a = 2 , n = 3 \ and \ a_3 = 26
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_3 =2+(3-1)d
\Rightarrow 26 -2=(2)d
\Rightarrow d = \frac{24}{2}= 12
Now,
a_2= a_1+d
a_2= 2+12 = 14
Therefore, the missing term is 14

Q3 (ii) In the following APs, find the missing terms in the boxes: \small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3

Answer:

Given AP series is
\small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3
Here, a_2 = 13 , n = 4 \ and \ a_4 = 3
Now,
a_2= a_1+d
a_1= a = 13 - d
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_4 =13-d+(4-1)d
\Rightarrow 3-13=-d+3d
\Rightarrow d = -\frac{10}{2}= -5
Now,
a_2= a_1+d
a_1= a = 13 - d= 13-(-5 ) = 18
And 
a_3=a_2+d
a_3=13-5 = 8
Therefore, missing terms are  18 and 8 
AP series is 18,13,8,3

Q3 (iii) In the following APs, find the missing terms in the boxes :   \small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}

Answer:

Given AP series is
\small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}
Here, a = 5 , n = 4 \ and \ a_4 = 9\frac{1}{2}= \frac{19}{2}
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_4 =5+(4-1)d
\Rightarrow \frac{19}{2} -5=3d
\Rightarrow d = \frac{19-10}{2\times 3} = \frac{9}{6} = \frac{3}{2}
Now,
a_2= a_1+d
a_2 = 5+\frac{3}{2} = \frac{13}{2}
And 
a_3=a_2+d
a_3=\frac{13}{2}+\frac{3}{2} = \frac{16}{2} = 8
Therefore, missing terms are  \frac{13}{2} and 8 
AP series is 5,\frac{13}{2}, 8 , \frac{19}{2}

Q3 (iv) In the following APs, find the missing terms in the boxes :   \small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6

Answer:

Given AP series is
\small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6
Here, a = -4 , n = 6 \ and \ a_6 = 6
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_6 =-4+(6-1)d
\Rightarrow 6+4 = 5d
\Rightarrow d = \frac{10}{5} = 2
Now,
a_2= a_1+d
a_2 = -4+2 = -2
And 
a_3=a_2+d
a_3=-2+2 = 0
And 
a_4 = a_3+d
a_4 = 0+2 = 2
And 
a_5 = a_4 + d
a_5 = 2+2 = 4
Therefore, missing terms are   -2,0,2,4 
AP series is -4,-2,0,2,4,6

Q3 (v) In the following APs, find the missing terms in the boxes :   \small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22

Answer:

Given AP series is
\small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22
Here, a_2 = 38 , n = 6 \ and \ a_6 = -22
Now,
a_2=a_1+d
a_1=a =38-d \ \ \ \ \ \ \ \ \ -(i)
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_6 =38-d+(6-1)d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow -22-38-=-d+5d
\Rightarrow d = -\frac{60}{4} = - 15
Now,
a_2= a_1+d
a_1 = 38-(-15) = 38+15 = 53
And 
a_3=a_2+d
a_3=38-15 = 23
And 
a_4 = a_3+d
a_4 = 23-15 = 8
And 
a_5 = a_4 + d
a_5 =8-15 = -7
Therefore, missing terms are   53,23,8,-7 
AP series is 53,38,23,8,-7,-22

Q4 Which term of the AP :  \small 3,8,13,18,...,  is \small 78 ?
Answer:

Given AP is 
\small 3,8,13,18,...,
Let suppose that nth term of AP is 78
Here, a = 3
And 
d = a_2-a_1 = 8 - 3 = 5
Now, we know that that
a_n = a + (n-1)d
\Rightarrow 78 = 3 + (n-1)5
\Rightarrow 78 -3 = 5n-5
\Rightarrow n = \frac{75 +5}{5}= \frac{80}{5} = 16
Therefore, value of  16th term of given AP is 78 

Q5 (i) Find the number of terms in each of the following APs :  \small 7,13,19,...,205

Answer:

Given AP series is 
\small 7,13,19,...,205
Let's suppose there are n terms in given AP
Then, 
a = 7 , a_n = 205
And 
d= a_2-a_1 = 13-7 = 6
Now, we know that
a_n =a + (n-1)d
\Rightarrow 205=7 + (n-1)6
\Rightarrow 205-7 = 6n-6
\Rightarrow n = \frac{198+6}{6} = \frac{204}{6} = 34
Therefore, there are 34 terms in given AP

Q5 (ii) Find the number of terms in each of the following APs :   \small 18,15\frac{1}{2},13,...,-47

Answer:

Given AP series is 
\small 18,15\frac{1}{2},13,...,-47
suppose there are n terms in given AP
Then, 
a = 18 , a_n = -47
And 
d= a_2-a_1 = \frac{31}{2}-18 = \frac{31-36}{2} = -\frac{5}{2}
Now, we know that
a_n =a + (n-1)d
\Rightarrow -47=18 + (n-1)\left ( -\frac{5}{2} \right )
\Rightarrow -47-18= -\frac{5n}{2}+\frac{5}{2}
\Rightarrow -\frac{5n}{2}= -65-\frac{5}{2}
\Rightarrow -\frac{5n}{2}= -\frac{135}{2}
\Rightarrow n = 27
Therefore, there are 27 terms in given AP

Q6 Check whether  \small -150  is a term of the AP :   \small 11,8,5,2...

Answer:

Given AP series is 
\small 11,8,5,2...
Here, a = 11
And 
d = a_2-a_1 = 8-11 = -3
Now, 
 suppose -150 is nth term of the given AP
Now, we know that
a_n = a+(n-1)d
\Rightarrow -150 = 11+(n-1)(-3)
\Rightarrow -150- 11=-3n+3
\Rightarrow =n = \frac{161+3}{3}= \frac{164}{3} = 54.66
Value of n is not an integer
Therefore, -150 is not a term of AP \small 11,8,5,2...

Q7 Find the  \small 31st term of an AP whose \small 11th term is \small 38 and the \small 16th term is \small 73.

Answer:

It is given that
 \small 11th term of an AP is  \small 38 and the \small 16th term is \small 73
Now,
a_{11} =38= a+ 10d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_{16} =73= a+ 15d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get 
a= -32 \ \ \ and \ \ \ d = 7
Now,
a_{31} = a+30d = -32 + 30\times 7 = -32+210 = 178
Therefore, 31st terms of given AP is 178

Q8 An AP consists of \small 50 terms of  which \small 3rd term is \small 12 and the last term is \small 106. Find the \small 29th term.
Answer:

It is given that
AP consists of \small 50 terms of  which \small 3rd term is \small 12 and the last term is \small 106
Now,
a_3 = 12=a+2d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_{50} = 106=a+49d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get 
a= 8 \ \ \ and \ \ \ d = 2
Now,
a_{29} = a+28d=8+28\times 2 = 8 +56 = 64
Therefore, 29th term of given AP is 64

Q9 If the \small 3rd and the \small 9th terms of an AP are \small 4  and  \small -8  respectively, which term of this AP is zero?
Answer:

It is given that
\small 3rd and the \small 9th terms of an AP are \small 4  and  \small -8  respectively
Now,
a_3 = 4=a+2d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_{9} = -8=a+8d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get 
a= 8 \ \ \ and \ \ \ d = -2
Now,
Let nth term of given AP is 0
Then,
a_{n} = a+(n-1)d
0 = 8+(n-1)(-2)
2n = 8+2= 10
n = \frac{10}{2} = 5
Therefore, 5th term of given AP is 0

Q10 The \small 17th term of an AP exceeds its \small 10th term by \small 7. Find the common difference.
Answer:

It is given that
\small 17th term of an AP exceeds its \small 10th term by \small 7
i.e.
a_{17}= a_{10}+7
\Rightarrow a+16d = a+9d+7
\Rightarrow a+16d - a-9d=7
\Rightarrow 7d=7
\Rightarrow d = 1
Therefore, the common difference of AP is 1

Q11 Which term of the AP : \small 3,15,27,39,...  will be \small 132 more than its  \small 54th  term?
Answer:

Given  AP is
\small 3,15,27,39,...
Here, a= 3
And
d= a_2-a_1 = 15 - 3 = 12
Now, let's suppose nth term of given AP is  \small 132 more than its  \small 54th  term
Then,
a_n= a_{54}+132
\Rightarrow a+(n-1)d = a+53d+132
\Rightarrow 3+(n-1)12 = 3+53\times 12+132
\Rightarrow 12n = 3+636+132+12
\Rightarrow 12n = 636+132+12
\Rightarrow n = \frac{780}{12}= 65
Therefore, 65th term of given AP is  \small 132 more than its  \small 54th  term

Q12 Two APs have the same common difference. The difference between their \small 100th terms is \small 100, what is the difference between their \small 1000th terms?
Answer:

It is given that
Two APs have the same common difference and  difference between their   \small 100th terms is  \small 100
i.e.
a_{100}-a'_{100}= 100
Let common difference of both the AP's is d 
\Rightarrow a+99d-a'-99d=100
\Rightarrow a-a'=100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - (i)
Now, difference between 1000th term is 
a_{1000}-a'_{1000}
\Rightarrow a+999d -a'-999d
\Rightarrow a-a'
\Rightarrow 100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i) )
Therefore, difference between 1000th term is 100

Q 13 How many three-digit numbers are divisible by \small 7?
Answer:

We know that the first three digit number divisible by 7 is 105 and last three-digit number divisible by 7 is 994
Therefore,
a = 105 , d = 7 \ and \ a_n = 994
Let there are n three digit numbers divisible by 7
Now, we know that
a_n = a+ (n-1)d
\Rightarrow 994 = 105 + (n-1)7
\Rightarrow 7n = 896
\Rightarrow n = \frac{896}{7} = 128
Therefore,  there are 128 three-digit numbers divisible by 7

Q14 How many multiples of \small 4 lie between \small 10  and \small 250?
Answer:

We know that the first number divisible by 4 between 10 to 250 is 12 and last  number divisible by 4 is 248
Therefore,
a = 12 , d = 4 \ and \ a_n = 248
Let there are n  numbers divisible by 4
Now, we know that
a_n = a+ (n-1)d
\Rightarrow 248 = 12 + (n-1)4
\Rightarrow 4n = 240
\Rightarrow n = \frac{240}{4} = 60
Therefore,  there are 60  numbers between 10 to 250 that are divisible by 4

Q15 For what value of \small n, are the \small nth terms of two APs:  \small 63,65,67,... and  \small 3,10,17,... equal?

Answer:

Given two AP's are
 \small 63,65,67,...   and    \small 3,10,17,...
Let first term and the common difference of two AP's are a , a' and d , d'
a = 63 \ , d = a_2-a_1 = 65-63 = 2
And
a' = 3 \ , d' = a'_2-a'_1 = 10-3 = 7
Now,
Let nth term of  both the AP's are equal
a_n = a'_n
\Rightarrow a+(n-1)d=a'+(n-1)d'
\Rightarrow 63+(n-1)2=3+(n-1)7
\Rightarrow 5n=65
\Rightarrow n=\frac{65}{5} = 13
Therefore, the 13th term of both the AP's are equal

Q16 Determine the AP whose third term is \small 16 and the \small 7th term exceeds the \small 5th term by \small 12.
Answer:

It is given that
3rd term of AP is \small 16 and the \small 7th term exceeds the \small 5th term by \small 12
i.e.
a_3=a+2d = 16 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And 
a_7=a_5+12
a+6d=a+4d+12
2d = 12
d = 6
Put the value of d in equation (i) we will get
a = 4
Now, AP with first term = 4 and common difference = 6 is 
4,10,16,22,.....

Q17 Find the \small 20th term from the last term of the AP :  \small 3,8,13,...,253.

Answer:

Given AP is 
\small 3,8,13,...,253
Here, a = 3 \ and \ a_n = 253
And
d = a_2-a_1=8-3 = 5
Let suppose there are n terms in the AP
Now, we know that
a_n = a+(n-1)d
253= 3+(n-1)5
5n = 255
n = 51
So, there are 51 terms in the given AP and 20th term from the last  will be 32th term from the starting
Therefore,
a_{32} = a+31d
a_{32} = 3+31\times 5 = 3+155 = 158
Therefore, 20th term from the of given AP is 158

Q18 The sum of the  \small 4th and  \small 8th terms of an AP is  \small 24 and the sum of the \small 6th and \small 10th terms is \small 44  Find the first three terms of the AP.
Answer:

It is given that
sum of the  \small 4th and  \small 8th terms of an AP is  \small 24 and the sum of the \small 6th and \small 10th terms is \small 44
i.e.
a_4+a_8=24
\Rightarrow a+3d+a+7d=24
\Rightarrow 2a+10d=24
\Rightarrow a+5d=12 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_6+a_{10}=44
\Rightarrow a+5d+a+9d=44
\Rightarrow 2a+14d=44
\Rightarrow a+7d=22 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get 
a= -13 \ and \ d= 5
Therefore,first three of AP with a = -13 and d = 5 is
-13,-8,-3

Q19 Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of  Rs 200 each year. In which year did his income reach Rs 7000?

Answer:

It is given that
Subba Rao started work at an annual salary of Rs 5000 and received an increment of  Rs 200 each year
Therefore, a = 5000 \ and \ d =200
Let's suppose after n years his salary will be Rs 7000
Now, we know that
a_n = a+(n-1)d
\Rightarrow 7000=5000+(n-1)200
\Rightarrow 2000=200n-200
\Rightarrow 200n=2200
\Rightarrow n = 11
Therefore, after 11years  his salary will be Rs 7000 
after 11 years, starting from 1995, his salary will reach to 7000, so we have to add 10 in 1995, because these numbers are in years
Thus , 1995+10 = 2005

Q20 Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by  Rs \small 1.75.  If in the \small nth week, her weekly savings become Rs \small 20.75, find \small n

Answer:

It is given that 
Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by  Rs \small 1.75
Therefore, a = 5 \ and \ d = 1.75
after  \small nth week, her weekly savings become Rs \small 20.75
Now, we know that 
a_n = a +(n-1)d
\Rightarrow 20.75= 5+(n-1)1.75
\Rightarrow 15.75= 1.75n-1.75
\Rightarrow 1.75n=17.5
\Rightarrow n=10
Therefore, after 10 weeks her saving will become Rs 20.75

NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 5.3

Q1 (i) Find the sum of the following APs:  \small 2,7,12,..., to \small 10 terms.

Answer:

Given AP is 
 \small 2,7,12,..., to \small 10 terms
Here, a = 2 \ and \ n = 10
And
d = a_2-a_1=7-2=5
Now, we know that 
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{10}{2}\left \{ 2\times 2 +(10-1)5\right \}
\Rightarrow S = 5\left \{ 4 +45\right \}
\Rightarrow S = 5\left \{ 49\right \}
\Rightarrow S =245
Therefore, the sum of AP  \small 2,7,12,..., to \small 10 terms is 245

Q1 (ii) Find the sum of the following APs:    \small -37,-33,-29,..., to \small 12 terms.

Answer:

Given AP is 
\small -37,-33,-29,..., to \small 12 terms.
Here, a = -37 \ and \ n = 12
And
d = a_2-a_1=-33-(-37)=4
Now, we know that 
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{12}{2}\left \{ 2\times (-37) +(12-1)4\right \}
\Rightarrow S = 6\left \{ -74 +44\right \}
\Rightarrow S = 5\left \{ -30\right \}
\Rightarrow S =-180
Therefore, the sum of AP  \small -37,-33,-29,..., to \small 12 terms. is -180

Q1 (iii) Find the sum of the following APs:   \small 0.6,1.7,2.8,..., to \small 100  terms.

Answer:

Given AP is 
 \small 0.6,1.7,2.8,..., to \small 100  terms..
Here, a = 0.6 \ and \ n = 100
And
d = a_2-a_1=1.7-0.6=1.1
Now, we know that 
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{100}{2}\left \{ 2\times (0.6) +(100-1)(1.1)\right \}
\Rightarrow S = 50\left \{ 1.2 +108.9\right \}
\Rightarrow S = 50\left \{ 110.1\right \}
\Rightarrow S =5505
Therefore, the sum of AP \small 0.6,1.7,2.8,..., to \small 100  terms. is 5505

Q1 (iv) Find the sum of the following APs:   \small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,  to   \small 11  terms.

Answer:

Given AP is 
 \small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,  to   \small 11  terms.
Here, a = \frac{1}{15} \ and \ n = 11
And
d = a_2-a_1=\frac{1}{12}-\frac{1}{15}= \frac{5-4}{60}= \frac{1}{60}
Now, we know that 
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{11}{2}\left \{ 2\times \frac{1}{15} +(11-1)(\frac{1}{60})\right \}
\Rightarrow S = \frac{11}{2}\left \{ \frac{2}{15} +\frac{1}{6}\right \}
\Rightarrow S = \frac{11}{2}\left \{ \frac{9}{30}\right \}
\Rightarrow S =\frac{99}{60}= \frac{33}{20}
Therefore, the sum of AP  \small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,  to   \small 11  terms. is \frac{33}{20}

 

Q2 (i) Find the sums given below :   \small 7+10\frac{1}{2}+14+...+84

Answer:

Given AP is 
\small 7+10\frac{1}{2}+14+...+84
We first need to find the number of terms
Here, a = 7 \ and \ a_n = 84
And
d = a_2-a_1=\frac{21}{2}-7= \frac{21-14}{2}= \frac{7}{2}
Let suppose there are n terms in the AP
Now, we know that
a_n = a+(n-1)d
\Rightarrow 84 = 7 + (n-1)\frac{7}{2}
\Rightarrow \frac{7n}{2}= 77+\frac{7}{2}
\Rightarrow n = 23
Now, we know that 
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{23}{2}\left \{ 2\times7 +(23-1)(\frac{7}{2})\right \}
\Rightarrow S = \frac{23}{2}\left \{ 14 +77\right \}
\Rightarrow S = \frac{23}{2}\left \{ 91\right \}
\Rightarrow S =\frac{2093}{2}=1046\frac{1}{2}
Therefore, the sum of AP \small 7+10\frac{1}{2}+14+...+84   is 1046\frac{1}{2}

Q2 (ii) Find the sums given below :   \small 34+32+30+...+10

Answer:

Given AP is 
\small 34+32+30+...+10
We first need to find the number of terms
Here, a = 34 \ and \ a_n = 10
And
d = a_2-a_1=32-34=-2
Let suppose there are n terms in the AP
Now, we know that
a_n = a+(n-1)d
\Rightarrow 10 = 34 + (n-1)(-2)
\Rightarrow -26 = -2n
\Rightarrow n = 13
Now, we know that 
S = \frac{n}{2}\left \{ a+a_n \right \}

\Rightarrow S = \frac{13}{2}\left \{ 44\right \}
\Rightarrow S =13\times 22 = 286
Therefore, the sum of AP \small 34+32+30+...+10   is 286

Q2 (iii) Find the sums given below :  \small -5+(-8)+(-11)+...+(-230)

Answer:

Given AP is 
\small -5+(-8)+(-11)+...+(-230)
We first need to find the number of terms
Here, a = -5 \ and \ a_n = -230
And
d = a_2-a_1=-8-(-5)= -3
Let suppose there are n terms in the AP
Now, we know that
a_n = a+(n-1)d
\Rightarrow -230 = -5 + (n-1)(-3)
\Rightarrow -228 = -3n
\Rightarrow n = 76
Now, we know that 
S = \frac{n}{2}\left \{ a+a_n \right \}
\Rightarrow S = \frac{76}{2}\left \{ (-5-230 )\right \}

\Rightarrow S = 38\left \{ -235\right \}
\Rightarrow S = -8930
Therefore, the sum of AP \small -5+(-8)+(-11)+...+(-230)   is -8930

Q3 (i) In an AP: given \small a=5, \small d=3,  \small a_n=50 , find \small n and \small S_n.

Answer:

It is given that
a = 5, d = 3 \ and \ a_n = 50
Let suppose there are n terms in the AP
Now, we know that
a_n = a+(n-1)d
\Rightarrow 50 = 5 + (n-1)3
\Rightarrow 48 = 3n
\Rightarrow n = 16
Now, we know that 
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{16}{2}\left \{ 2\times(5) +(16-1)(3)\right \}
\Rightarrow S = 8\left \{ 10+45\right \}
\Rightarrow S = 8\left \{ 55\right \}
\Rightarrow S =440
Therefore, the sum of the given AP  is 440

Q3 (ii) In an AP: given \small a=7\small a_1_3=35, find \small d and \small S_1_3.

Answer:

It is given that
a = 7 \ and \ a_{13} = 35
a_{13}= a+12d = 35
        = 12d = 35-7 = 28
d = \frac{28}{12}= \frac{7}{3}
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{ 2\times(7) +(13-1)(\frac{7}{3})\right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{14 +28\right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{42\right \}
\Rightarrow S_{13} = 13 \times 21 = 273
Therefore, the sum of given AP  is 273

Q3 (iii) In an AP: given  \small a_1_2=37,d=3, find \small a and \small S_1_2.

Answer:

It is given that
d = 3 \ and \ a_{12} = 37
a_{12}= a+11d = 37
        = a= 37-11\times 3 = 37-33=4
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{12} = \frac{12}{2}\left \{ 2\times(4) +(12-1)3\right \}
\Rightarrow S_{12} = 6\left \{ 8+33\right \}
\Rightarrow S_{12} = 6\left \{41\right \}
\Rightarrow S_{12} =246
Therefore, the sum of given AP  is 246

Q3 (iv) In an AP: given \small a_3=15, S_1_0=125, find \small d and \small S_1_0
Answer:

It is given that
\small a_3=15, S_1_0=125
a_{3}= a+2d = 15 \ \ \ \ \ \ \ \ -(i)
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{10} = \frac{10}{2}\left \{ 2\times(a) +(10-1)d\right \}
\Rightarrow 125 = 5\left \{ 2a+9d\right \}
\Rightarrow 2a+9d = 25 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get 
a= 17 \ and \ d = -1
Now,
a_{10} = a+ 9d = 17 + 9(-1)= 17-9 = 8
Therefore, the value of d and 10th terms is -1 and 8 respectively

Q3 (v) In an AP: given \small d=5, S_9=75, find \small a and \small a_9.

Answer:

It is given that
\small d=5, S_9=75
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{9} = \frac{9}{2}\left \{ 2\times(a) +(9-1)5\right \}
\Rightarrow 75= \frac{9}{2}\left \{ 2a +40\right \}
\Rightarrow 150= 18a+360
\Rightarrow a = -\frac{210}{18}=-\frac{35}{3}
Now,
a_{9} = a+ 8d = -\frac{35}{3} + 8(5)= -\frac{35}{3}+40 = \frac{-35+120}{3}= \frac{85}{3}

Q3 (vi) In an AP: given  \small a=2,d=8,S_n=90, find \small n and \small a_n.

Answer:

It is given that
\small a=2,d=8,S_n=90,
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 90 = \frac{n}{2}\left \{ 2\times(2) +(n-1)8\right \}
\Rightarrow 180 = n\left \{ 4+8n-8\right \}
\Rightarrow 8n^2-4n-180=0
\Rightarrow 4(2n^2-n-45)=0
\Rightarrow 2n^2-n-45=0
\Rightarrow 2n^2-10n+9n-45=0
\Rightarrow (n-5)(2n+9)=0
\Rightarrow n = 5 \ \ and \ \ n = - \frac{9}{2}
n can not be negative so the only the value of n is 5
Now,
a_{5} = a+ 4d = 2+4\times 8 = 2+32 = 34
Therefore, value of n and nth term is 5 and 34 respectively

Q3 (vii) In an AP: given  \small a=8,a_n=62,S_n=210, find \small n and \small d.

Answer:

It is given that
\small a=8,a_n=62,S_n=210,
Now, we know that
a_n = a+(n-1)d
62 = 8+(n-1)d
(n-1)d= 54 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 210 = \frac{n}{2}\left \{ 2\times(8) +(n-1)d\right \}
\Rightarrow 420 = n\left \{ 16+54 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow 420 = n\left \{ 70 \right \}
\Rightarrow n = 6
Now, put this value in (i) we will get
d = \frac{54}{5}
Therefore, value of n and d are 6 \ and \ \frac{54}{5} respectively

Q3 (viii) In an AP: given \small a_n=4,d=2,S_n=-14,  find \small n and \small a .

Answer:

It is given that
\small a_n=4,d=2,S_n=-14,
Now, we know that
a_n = a+(n-1)d
4 = a+(n-1)2
a+2n = 6\Rightarrow a = 6-2n \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow -14 = \frac{n}{2}\left \{ 2\times(a) +(n-1)2\right \}
\Rightarrow -28 = n\left \{ 2(6-2n)+2n-2 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow -28 = n\left \{ 10-2n \right \}
\Rightarrow 2n^2-10n-28=0
\Rightarrow 2(n^2-5n-14)=0
\Rightarrow n^2-7n+2n-14=0
\Rightarrow(n+2)(n-7)=0
\Rightarrow n = -2 \ \ and \ \ n = 7
Value of n cannot be negative so the only the value of n is 7
Now, put this value in (i) we will get
a = -8
Therefore, the value of n and a are 7 and -8  respectively

Q3 (ix) In an AP: given  \small a=3,n=8,S=192,  find \small d .

Answer:

It is given that
\small a=3,n=8,S=192,
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 192 = \frac{8}{2}\left \{ 2\times(3) +(8-1)d\right \}
\Rightarrow 192=4\left \{6 +7d\right \}
\Rightarrow 7d = 48-6
\Rightarrow d = \frac{42}{7} = 6
Therefore, the value of d is 6

Q3 (x) In an AP: given  \small l=28,S=144 \ and \ n = 9  and there are total \small 9 terms. Find \small a.

Answer:

It is given that
\small l=28,S=144 \ and \ n = 9
Now, we know that
l = a_n = a+(n-1)d
28 = a_n = a+(n-1)d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 144 = \frac{9}{2}\left \{ a + a +(n-1)d\right \}
\Rightarrow 288 =9\left \{ a+ 28\right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(using \ (i))
\Rightarrow a+28= 32
\Rightarrow a=4
Therefore, the value of a is 4

Q4 How many terms of the AP:  \small 9,17,25,... must be taken to give a sum of \small 636?

Answer:

Given AP is 
\small 9,17,25,... 
Here, a =9 \ and \ d = 8
And  S_n = 636
Now , we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 636 = \frac{n}{2}\left \{ 18+(n-1)8 \right \}
\Rightarrow 1272 = n\left \{ 10+8n \right \}
\Rightarrow 8n^2+10n-1272=0
\Rightarrow 2(4n^2+5n-636)=0
\Rightarrow 4n^2+53n-48n-636=0
\Rightarrow (n-12)(4n+53)=0
\Rightarrow n = 12 \ \ and \ \ n = - \frac{53}{4}
Value of  n  can not be negative so the only the value of n is 12
Therefore, the sum of 12 terms of AP  \small 9,17,25,... must be taken to give a sum of \small 636.

Q5 The first term of an AP is \small 5, the last term is \small 45 and the sum is \small 400. Find the number of terms and the common difference.

Answer:

It is given that
\small a=5,a_n=45,S_n=400,
Now, we know that
a_n = a+(n-1)d
45 = 5+(n-1)d
(n-1)d= 40 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 400 = \frac{n}{2}\left \{ 2\times(5) +(n-1)d\right \}
\Rightarrow 800 = n\left \{ 10+40 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow 800 = n\left \{ 50 \right \}
\Rightarrow n = 16
Now, put this value in (i) we will get
d = \frac{40}{15}= \frac{8}{3}
Therefore, value of n and d are 16 \ and \ \frac{8}{3} respectively

Q6 The first and the last terms of an AP are  \small 17 and  \small 350  respectively. If the common difference is \small 9, how many terms are there and what is their sum?

Answer:

It is given that
\small a=17,l=350,d=9,
Now, we know that
a_n = a+(n-1)d
350 = 17+(n-1)9
(n-1)9 = 333
(n-1)=37
n = 38

Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{38}= \frac{38}{2}\left \{ 2\times(17) +(38-1)9\right \}
\Rightarrow S_{38}= 19\left \{ 34 +333\right \}
\Rightarrow S_{38}= 19\left \{367\right \}
\Rightarrow S_{38}= 6973
Therefore, there are 38 terms  and their sun is 6973

Q7 Find the sum of first \small 22 terms of an AP in which \small d=7 and \small 22nd term is \small 149.

Answer:

It is given that
\small a_{22}=149,d=7,n = 22
Now, we know that
a_{22} = a+21d
149 = a+21\times 7
a = 149 - 147 = 2
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{22}= \frac{22}{2}\left \{ 2\times(2) +(22-1)7\right \}
\Rightarrow S_{22}= 11\left \{ 4 +147\right \}
\Rightarrow S_{22}= 11\left \{ 151\right \}
\Rightarrow S_{22}= 1661
Therefore, there are 22 terms  and their sum is 1661

Q8 Find the sum of first \small 51 terms of an AP whose second and third terms are \small 14 and \small 18 respectively.

Answer:

It is given that
\small a_{2}=14,a_3=18,n = 51
And d= a_3-a_2= 18-14=4
Now,
a_2 = a+d
a= 14-4 = 10
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{51}= \frac{51}{2}\left \{ 2\times(10) +(51-1)4\right \}
\Rightarrow S_{51}= \frac{51}{2}\left \{ 20 +200\right \}
\Rightarrow S_{51}= \frac{51}{2}\left \{ 220\right \}
\Rightarrow S_{51}= 51 \times 110
\Rightarrow S_{51}=5304
Therefore, there are 51 terms  and their sum is 5610

Q9 If the sum of first \small 7 terms of an AP is \small 49 and that of \small 17 terms is \small 289 , find the sum of first \small n terms.

Answer:

It is given that
S_7 = 49 \ and \ S_{17}= 289
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{7}= \frac{7}{2}\left \{ 2\times(a) +(7-1)d\right \}
\Rightarrow 98= 7\left \{ 2a +6d\right \}
\Rightarrow a +3d = 7 \ \ \ \ \ \ \ -(i)
Similarly,
\Rightarrow S_{17}= \frac{17}{2}\left \{ 2\times(a) +(17-1)d\right \}
\Rightarrow 578= 17\left \{ 2a +16d\right \}
\Rightarrow a +8d = 17 \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get 
a = 1 and d = 2
Now, the sum of first n terms is 
S_n = \frac{n}{2}\left \{ 2\times 1 +(n-1)2 \right \}
S_n = \frac{n}{2}\left \{ 2 +2n-2 \right \}
S_n = n^2
Therefore, the sum of n terms  is  n^2

Q10 (i) Show that  \small a_1,a_2,...,a_n,...  form an AP where an is defined as below :   \small a_n=3+4n Also find the sum of the first \small 15 terms.

Answer:

It is given that 
\small a_n=3+4n
We will check values of a_n for different values of n
a_1 = 3+4(1) =3+4= 7
a_2 = 3+4(2) =3+8= 11
a_3 = 3+4(3) =3+12= 15
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to and common difference (d) equals to 4
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(7) +(15-1)4\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 14 +56\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 70\right \}
\Rightarrow S_{15}= 15 \times 35
\Rightarrow S_{15}= 525
Therefore, the sum of 15 terms  is  525

Q10 (ii) Show that  \small a_1,a _2,...,a_n,... form an AP where an is defined as below : \small a_n=9-5n .Also find the sum of the first \small 15 terms in each case.

Answer:

It is given that 
\small a_n=9-5n
We will check values of a_n for different values of n
a_1 = 9-5(1) =9-5= 4
a_2 = 9-5(2) =9-10= -1
a_3 = 9-5(3) =9-15= -6
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 4 and common difference (d) equals to -5
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(4) +(15-1)(-5)\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 8 -70\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ -62\right \}
\Rightarrow S_{15}= 15 \times (-31)
\Rightarrow S_{15}= -465
Therefore, the sum of 15 terms is -465

Q11 If the sum of the first \small n terms of an AP is  \small 4n-n^2, what is the first term (that is  \small S_1)? What is the sum of first two terms? What is the second term? Similarly, find the \small 3rd, the \small 10th and the \small nth terms

Answer:

It is given that 
the sum of the first \small n terms of an AP is  \small 4n-n^2
Now,
\Rightarrow S_n = 4n-n^2
Now, first term is 
\Rightarrow S_1 = 4(1)-1^2=4-1=3
Therefore, first term is 3
Similarly,
\Rightarrow S_2 = 4(2)-2^2=8-4=4
Therefore, sum of first two terms is 4
Now, we know that
\Rightarrow S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_2 = \frac{2}{2}\left \{ 2\times 3+(2-1)d \right \}
\Rightarrow 4 = \left \{6+d \right \}
\Rightarrow d = -2
Now,
a_2= a+d = 3+(-2 )= 1
Similarly,
a_3= a+2d = 3+2(-2 )= 3-4=-1
a_{10}= a+9d = 3+9(-2 )= 3-18=-15
a_{n}= a+(n-1)d = 3+(n-1)(-2 )= 5-2n

Q12 Find the sum of the first \small 40 positive integers divisible by \small 6.

Answer:

Positive integers divisible by 6 are 
6,12,18,...
This is an AP with 
here, \ a = 6 \ and \ d = 6
Now, we know that 
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{40}= \frac{40}{2}\left \{ 2\times 6+(40-1)6 \right \}
\Rightarrow S_{40}= 20\left \{12+234 \right \}
\Rightarrow S_{40}= 20\left \{246 \right \}
\Rightarrow S_{40}= 4920
Therefore,  sum of the first \small 40 positive integers divisible by  \small 6 is 4920

Q13 Find the sum of the first \small 15  multiples of  \small 8.

Answer:

First 15 multiples of 8 are  
8,16,24,...
This is an AP with 
here, \ a = 8 \ and \ d = 8
Now, we know that 
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times 8+(15-1)8 \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 16+112 \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 128 \right \}
\Rightarrow S_{15}= 15 \times 64 = 960
Therefore,  sum of the first 15 multiple of 8 is 960

Q14 Find the sum of the odd numbers between \small 0 and \small 50.

Answer:

The odd number between 0 and 50 are  
1,3,5,...49
This is an AP with 
here, \ a = 1 \ and \ d = 2
There are total 25 odd number between 0 and 50
Now, we know that 
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{25}= \frac{25}{2}\left \{ 2\times 1+(25-1)2 \right \}
\Rightarrow S_{25}= \frac{25}{2}\left \{ 2+48 \right \}
\Rightarrow S_{25}= \frac{25}{2}\times 50
\Rightarrow S_{25}= 25 \times 25 = 625
Therefore,  sum of the odd numbers between \small 0 and \small 50  625

Q15 A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs \small 200  for the first day, Rs \small 250 for the second day, Rs \small 300 for the third day, etc., the penalty for each succeeding day being Rs \small 50 more than for the preceding day. How much money the contractor has to pay a penalty, if he has delayed the work by \small 30 days?

Answer:

It is given that
Penalty for delay of completion beyond a certain date is Rs \small 200  for the first day, Rs \small 250 for the second day, Rs \small 300 for the third day and  penalty for each succeeding day being Rs \small 50 more than for the preceding day
We can clearly see that
200,250,300,..... is an AP  with 
a = 200 \ and \ d = 50
Now, the penalty for 30 days is given by the expression
S_{30}= \frac{30}{2}\left \{ 2\times 200+(30-1)50 \right \}
S_{30}= 15\left ( 400+1450 \right )
S_{30}= 15 \times 1850
S_{30}= 27750
Therefore, the penalty for 30 days is 27750

Q16 A sum of Rs \small 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs \small 20 less than its preceding prize, find the value of each of the prizes.

Answer:

It is given that
Each price is decreased by 20 rupees, 
Therefore,  d = -20 and there are total 7 prizes so n = 7 and sum of prize money is Rs 700 so S_7 = 700
Let a be the prize money given to the 1st student
Then,
S_7 = \frac{7}{2}\left \{ 2a+(7-1)(-20) \right \}
700 = \frac{7}{2}\left \{ 2a-120 \right \}
2a - 120 = 200
a = \frac{320}{2}= 160
Therefore, the prize given to the first student is Rs 160
Now,
Let a_2,a_2,...,a_7 is the prize money given to the next 6 students
then,
a_2 = a+d = 160+(-20)=160-20=140
a_3 = a+2d = 160+2(-20)=160-40=120  
a_4 = a+3d = 160+3(-20)=160-60=100
a_5 = a+4d = 160+4(-20)=160-80=80
a_6 = a+5d = 160+5(-20)=160-100=60
a_7 = a+6d = 160+6(-20)=160-120=40
Therefore, prize money given to 1 to 7 student is 160,140,120,100,80,60.40

Q17 In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I???? will plant \small 1 tree, a section of Class II will plant \small 2 trees and so on till Class XII. There are three sections in each class. How many trees will be planted by the students?

Answer:

First there are 12 classes and each class has 3 sections
Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1.Thus every class will plant 3 times the number of their class 
Similarly,

No. of trees planted by 3 sections of class 1 = 3

No. of trees planted by 3 sections of class 2 = 6

No. of trees planted by 3 sections of class 3 = 9

No. of trees planted by 3 sections of class 4 = 12
Its clearly an AP with first term(a)3 and common difference(d)3 and  total number of classes(n) = 12

Now, number of trees planted by 12 classes is given by
S_{12}= \frac{12}{2}\left \{ 2\times 3+(12-1)\times 3 \right \}
S_{12}= 6\left ( 6+33 \right )
S_{12}= 6 \times 39 = 234
Therefore,  number of trees planted by 12 classes is 234

Q18 A spiral is made up of successive semicircles, with centres alternately at \small A and \small B??????, starting with centre at \small A, of radii \small 0.5\hspace {1mm}cm,1.0\hspace {1mm}cm,1.5\hspace {1mm}cm,2.0\hspace {1mm}cm,...  as shown in Fig. \small 5.4. What is the total length of such a spiral made up of thirteen  consecutive semicircles? (Take   \pi =\frac{22}{7}  )

               

[Hint : Length of successive semicircles is  \small l_1,l_2,l_3,l_4,...  with centres at  \small A,B,A,B,..., respectively.]

Answer:

From the above-given figure

Circumference of 1st semicircle l_1 = \pi r_1 = 0.5\pi

Similarly,

Circumference of 2nd semicircle l_2 = \pi r_2 = \pi

Circumference of 3rd semicircle l_3 = \pi r_3 = 1.5\pi

It is clear that this is an AP with a = 0.5\pi \ and \ d = 0.5\pi

Now, sum of length of 13 such semicircles is given by

S_{13} = \frac{13}{2}\left \{ 2\times 0.5\pi + (13-1)0.5\pi\right \}
S_{13} = \frac{13}{2}\left ( \pi+6\pi \right )
S_{13} = \frac{13}{2}\times 7\pi
S_{13} = \frac{91\pi}{2} = \frac{91}{2}\times \frac{22}{7}=143
Therefore, sum of length of 13 such semicircles is 143 cm 

Q19 \small 200  logs are stacked in the following manner: \small 20 logs in the bottom row, \small 19  in the next row, \small 18 in the row next to it and so on (see Fig. \small 5.5). In how many rows are the \small 200 logs placed and how many logs are in the top row?

               

Answer:

As the rows are going up, the no of logs are decreasing, 
We can clearly see that 20, 19, 18, ..., is an AP.
and here  a = 20 \ and \ d = -1 
Let suppose 200 logs are arranged in 'n' rows,
Then, 
S_n = \frac{n}{2}\left \{ 2\times 20 +(n-1)(-1) \right \}
200 = \frac{n}{2}\left \{ 41-n \right \}
\Rightarrow n^2-41n +400 = 0
\Rightarrow n^2-16n-25n +400 = 0
\Rightarrow (n-16)(n-25) = 0
\Rightarrow n = 16 \ \ and \ \ n = 25
Now,
case (i) n = 25
a_{25} =a+24d = 20+24\times (-1)= 20-24=-4
But number of rows can not be in negative numbers 
Therefore, we will reject the value n = 25

case (ii) n = 16

a_{16} =a+15d = 20+15\times (-1)= 20-15=5
Therefore, the number of rows in which 200 logs are arranged is equal to 5

Q20 In a potato race, a bucket is placed at the starting point, which is \small 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. \small 5.6).

              

A competitor starts from the bucket, picks up the nearest potato, runs back with it,  drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What  is the total distance the competitor has to run?

            [Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is  \small 2\times5+2\times (5+3)  ]

Answer:

Distance travelled by the competitor in picking and dropping 1st potato = 2 \times 5 = 10 \ m

Distance travelled by the competitor in picking and dropping 2nd potato = 2 \times (5+3) =2\times 8 = 16 \ m

Distance travelled by the competitor in picking and dropping 3rd potato = 2 \times (5+3+3) =2\times 11 = 22 \ m

and so on
we can clearly see that it is an AP with first term (a) = 10  and common difference(d) = 6
There are 10 potatoes in the line 
Therefore, total distance travelled by the competitor in picking and dropping  potatoes is 
S_{10}= \frac{10}{2}\left \{ 2\times 10+(10-1)6 \right \}
S_{10}= 5\left ( 20+54 \right )
S_{10}= 5\times 74 = 370

Therefore, the total distance travelled by the competitor in picking and dropping  potatoes is  370 m

 

NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 5.4

Q1 Which term of the AP: is its first negative term?  [Hint : Find n for a_n<0 ] 

Answer:

Given AP is 
\small 121,117,113,...,
Here a = 121 \ and \ d = -4
Let suppose nth term of the AP is first negative term 
Then,
a_n = a+ (n-1)d
If nth term is negative then a_n < 0
\Rightarrow 121+(n-1)(-4) < 0
\Rightarrow 125<4n
\Rightarrow n > \frac{125}{4}=31.25
Therefore, first negative term must be 32nd term

Q2 The sum of the third and the seventh terms of an AP is \small 6 and their product is \small 8. Find the sum of first sixteen terms of the AP.

Answer:

It is given that sum of third and seventh terms of an AP are and their product is \small 8
a_3= a+ 2d
a_7= a+ 6d
Now,
a_3+a_7= a+ 2d+a+6d= 6
\Rightarrow 2a+8d = 6
\Rightarrow a+4d = 3 \Rightarrow a = 3-4d \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And 
a_3.a_7 = (a+2d).(a+6d)=a^2+8ad +12d^2 = 8 \ \ \ \ \ \ \ -(ii)
put value from equation (i) in (ii) we will get 
\Rightarrow (3-4d)^2+8(3-4d)d+12d^2= 8
\Rightarrow 9+16d^2-24d+24d-32d^2+12d^2=8
\Rightarrow 4d^2 = 1
\Rightarrow d = \pm \frac{1}{2}
Now,
case (i)  d = \frac{1}{2} 

a= 3 - 4 \times \frac{1}{2} = 1
Then,
S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\frac{1}{2} \right \}

S_{16}=76

case (ii)  d = -\frac{1}{2}
a= 3 - 4 \times \left ( -\frac{1}{2} \right ) = 5

Then,
S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\left ( -\frac{1}{2} \right ) \right \}

S_{16}=20

Q3 A ladder has rungs \small 25 cm apart. (see Fig. \small 5.7). The rungs decrease uniformly in length from \small 45 cm at the bottom to \small 25 cm at the top. If the top and the bottom rungs are  \small 2\frac{1}{2}  m apart, what is the length of the wood required for the rungs? [Hint: Number of rungs  =\frac{250}{25}+1]

                    

 

Answer:

It is given that
The total distance between the top and bottom rung = 2\frac{1}{2}\ m = 250cm
Distance between any two rungs = 25 cm
Total number of rungs = \frac{250}{25}+1= 11 
And it is also given that bottom-most rungs is of 45 cm length and topmost is of 25 cm length.As it is given that the length of rungs decrease uniformly, it will form an AP with a = 25 , a_{11} = 45 \ and \ n = 11
Now, we know that 
a_{11}= a+ 10d

\Rightarrow 45=25+10d
\Rightarrow d = 2
Now, total  length of the wood required for the rungs is equal to 
S_{11} = \frac{11}{2}\left \{ 2\times 25+(11-1)2 \right \}
S_{11} = \frac{11}{2}\left \{ 50+20 \right \}
S_{11} = \frac{11}{2}\times 70
S_{11} =385 \ cm
Therefore, the total  length of the wood required for the rungs is equal to  385 cm

Q4 The houses of a row are numbered consecutively from \small 1 to \small 49. Show that there is a value of  \small x such that the sum of the numbers of the houses preceding the house numbered \small x is equal to the sum of the numbers of the houses following it. Find this value of \small x. [Hint : \small S_{x-1}=S_4_9-S_x]

Answer:

It is given that the sum of the numbers of the houses preceding the house numbered \small x is equal to the  sum of the numbers of the houses following it
And 1,2,3,.....,49 form an AP with a = 1 and d = 1
Now, we know that
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
Suppose their exist an n term such that ( n < 49)
Now, according to given conditions 
Sum of first n - 1 terms of AP = Sum of terms following the nth term
Sum of first n - 1 term of AP = Sum of whole AP - Sum of first m terms of AP
i.e.
S_{n-1}=S_{49}-S_n

\frac{n-1}{2}\left \{ 2a+((n-1)-1)d \right \}=\frac{49}{2}\left \{ 2a+(49-1)d \right \}-\frac{n}{2}\left \{ 2a+(n-1)d \right \}

\frac{n-1}{2}\left \{ 2+(n-2) \right \}=\frac{49}{2}\left \{ 2+48 \right \}-\frac{n}{2}\left \{ 2+(n-1) \right \}

\frac{n-1}{2}\left \{ n\right \}=\frac{49}{2}\left \{ 50 \right \}-\frac{n}{2}\left \{ n+1 \right \}

\frac{n^2}{2}-\frac{n}{2}=1225-\frac{n^2}{2}-\frac{n}{2}

n^2 = 1225
n = \pm 35

Given House number are not  negative so we reject n = -35

Therefore, the sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 35

Q5 A small terrace at a football ground comprises of  \small 15 steps each of which is \small 50 m long and   built  of solid concrete. Each step has a rise of   \small \frac{1}{4}\: m   and a tread of   \small \frac{1}{2}\: m . (see Fig. \small 5.8).  Calculate the total volume of concrete required to build the terrace.                               
 [Hint: Volume of concrete required to build the first step  \small =\frac{1}{4}\times \frac{1}{2}\times 50\: m^3  ]

                

 

Answer:

It is given that 
football ground comprises of  \small 15 steps each of which is \small 50 m long and Each step has a rise of   \small \frac{1}{4}\: m   and a tread of   \small \frac{1}{2}\: m
Now,
The volume required to make the first step = \frac{1}{4}\times \frac{1}{2}\times 50 = 6.25 \ m^3

Similarly,

The volume required to make 2nd step = \left ( \frac{1}{4}+\frac{1}{4}\right )\times \frac{1}{2}\times 50=\frac{1}{2}\times \frac{1}{2}\times 50 = 12.5 \ m^3
And
The volume required to make 3rd step = \left ( \frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right )\times \frac{1}{2}\times 50=\frac{3}{4}\times \frac{1}{2}\times 50 = 18.75 \ m^3

And so on
We can clearly see that this is an AP with a= 6.25 \ and \ d = 6.25
Now, the total volume of concrete required to build the terrace of 15 such step is 
S_{15} =\frac{15}{2}\left \{ 2 \times 6.25 +(15-1)6.25 \right \}
S_{15} =\frac{15}{2}\left \{ 12.5 +87.5\right \}
S_{15} =\frac{15}{2}\times 100
S_{15} =15\times 50 = 750
Therefore, the total volume of concrete required to build the terrace of 15 such steps is 750 \ m^3

NCERT solutions for class 10 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers

Chapter 2

NCERT solutions for class 10 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables

Chapter 4

CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations

Chapter 5

NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions

Chapter 6

Solutions of NCERT class 10 maths chapter 6 Triangles

Chapter 7

CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry

Chapter 8

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

Chapter 9

Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry

Chapter 10

CBSE NCERT solutions class 10 maths chapter 10 Circles

Chapter 11

NCERT solutions  for class 10 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles

Chapter 13

CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes

Chapter 14

NCERT solutions for class 10 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 10 maths chapter 15 Probability

Solutions of NCERT class 10 subject wise

NCERT solutions for class 10 maths

Solutions of NCERT for class 10 science

How to use NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions?

  • First of all, go through the conceptual theory given in the NCERT textbook.

  • After going through the conceptual part, jump on to practice exercises available.

  • While solving the exercises if you are facing problems in any specific question, then take the help of NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions.

  • Once you have done the practice exercises you can move to previous year questions.

Keep working hard & happy learning!

 

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