NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

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 Rs 15 for the first km and Rs 8 for each additional km.

Answer:

It is given that

Fare for 1st km= Rs. 15
And after that Rs 8 for each additional km
Now,
Fare for 2nd km= Fare of first km+ Additional fare for 1 km

= Rs. 15+8=Rs23

Fare for 3rd km= Fare of first km+ Fare of additional second km+ Fare of additional third km

= Rs. 23+8= Rs 31

Fare of nkm=15+8×(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 1/4 of the air remaining in the cylinder at a time.

Answer:

It is given that

vacum pump removes 14 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

=1−14=34

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

=34×14=316

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

=34−316=12−316=916

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 Rs 150 for the first metre and rises by Rs 50 for each subsequent meter.

Answer:

It is given that
Cost of digging of 1st meter = Rs 150
and
rises by Rs 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 Rs 10000 is deposited at compound interest at 8 % per annum .

Answer:

Amount in the beginning = Rs. 10000

Interest at the end of 1 st year at the rate of 8%

is 8% of 10000=8×10000100=800

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

=10000+800=10800

Now,
Interest at the end of 2 nd year at rate of 8%

is 8% of 10800=8×10800100=864

Therefore,, amount at the end of 2 ^nd year

= 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 a = 10, d = 10

Answer:

It is given that

a=10,d=10

Now,

a1=a=10a2=a1+d=10+10=20a3=a2+d=20+10=30a4=a3+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: a=-2, d=0
Answer:

It is given that

a=−2,d=0

Now,

a1=a=−2a2=a1+d=−2+0=−2a3=a2+d=−2+0=−2a4=a3+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 a=4, d=-3

Answer:

It is given that

a=4,d=−3

Now,

a1=a=4a2=a1+d=4−3=1a3=a2+d=1−3=−2a4=a3+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 a=−1,d=12

Answer:

It is given that

a=−1,d=12

Now,

a1=a=−1a2=a1+d=−1+12=−12a3=a2+d=−12+12=0a4=a3+d=0+12=12

Therefore, the first four terms of the given series are −1,−12,0,12

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

Answer:

It is given that

a=−1.25,d=−0.25

Now,

a1=a=−1.25a2=a1+d=−1.25−0.25=−1.50a3=a2+d=−1.50−0.25=−1.75a4=a3+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: 3,1,−1,−3,…

Answer:

Given AP series is

3,1,−1,−3,…

Now, first term of this AP series is 3
Therefore,
First-term of AP series (a)=3
Now,

a1=3 and a2=1

And common difference (d)=a2−a1=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: −5,−1,3,7,…

Answer:

Given AP series is

−5,−1,3,7,…

Now, the first term of this AP series is −5
Therefore,
First-term of AP series (a)=−5
Now,

a1=−5 and a2=−1

And common difference (d)=a2−a1=−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: 13,53,93,133,…

Answer:

Given AP series is

13,53,93,133,…
Now, the first term of this AP series is 13
Therefore,
Thle first term of AP series (a)=13

Now,

a1=13 and a2=53

And common difference (d) =a2−a1=53−13=5−13=43
Therefore, the first term and the common difference is 13 and 43 respectively

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

Answer:

Given AP series is

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,

a1=0.6 and a2=1.7

And common difference (d)=a2−a1=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. 2, 4, 8, 12...

Answer:

Given series is

2,4,8,12,…

Now, the first term to this series is = 2

Now,

a1=2 and a2=4 and a3=8a2−a1=4−2=2a3−a2=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. 2,52,3,72,…

Answer:

Given series is

2,52,3,72,…

Now,
first term to this series is =2
Now,

a1=2 and a2=52 and a3=3 and a4=72a2−a1=52−2=5−42=12a3−a2=3−52=6−52=12a4−a3=72−3=7−62=12

We can clearly see that the difference between terms are equal and equal to 12 Hence, given series is in AP
Now, the next three terms are

a5=a4+d=72+12=82=4

a6=a5+d=4+12=8+12=92a7=a6+d=92+12=102=5

Therefore, next three terms of given series are 4,92,5

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

Answer:

Given series is

−1.2,−3.2,−5.2,−7.2,…

Now,
the first term to this series is =−1.2
Now,

a1=−1.2 and a2=−3.2 and a3=−5.2 and a4=−7.2a2−a1=−3.2−(−1.2)=−3.2+1.2=−2a3−a2=−5.2−(−3.2)=−5.2+3.2=−2a4−a3=−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

a5=a4+d=−7.2−2=−9.2a6=a5+d=−9.2−2=−11.2a7=a6+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. −10,−6,−2,2,…

Answer:

Given series is

−10,−6,−2,2,…

Now,
the first term to this series is =−10
Now,

a1=−10 and a2=−6 and a3=−2 and a4=2a2−a1=−6−(−10)=−6+10=4a3−a2=−2−(−6)=−2+6=4a4−a3=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

a5=a4+d=2+4=6a6=a5+d=6+4=10a7=a6+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. 3,3+2,3+22,3+32,…

Answer:

Given series is

3,3+2,3+22,3+32,…

Now,
the first term to this series is =3
Now,

a1=3 and a2=3+2 and a3=3+22 and a4=3+32a2−a1=3+2−3=2a3−a2=3+22−3−2=2a4−a3=3+32−3−22=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

a5=a4+d=3+32+2=3+42a6=a5+d=3+42+2=3+52a7=a6+d=3+52+2=3+62

Therefore, next three terms of given series are 3+42,3+52,3+62

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

Answer:

Given series is

0.2,0.22,0.222,0.2222,…

Now, the first term to this series is =0.2
Now,

a1=0.2 and a2=0.22 and a3=0.222 and a4=0.2222a2−a1=0.22−0.2=0.02a3−a2=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. 0,−4,−8,−12,…

Answer:

Given series is

Now,
first term to this series is = 0

0,−4,−8,−12,…

Now,
first term to this series is =0
Now,

a1=0 and a2=−4 and a3=−8 and a4=−12a2−a1=−4−0=−4a3−a2=−8−(−4)=−8+4=−4a4−a3=−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

a5=a4+d=−12−4=−16a6=a5+d=−16−4=−20a7=a6+d=−20−4=−24

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



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. −12,−12,−12,−12,…

Answer:

Given series is

−12,−12,−12,−12,…

Now, the first term to this series is =−12
Now,

a1=−12 and a2=−12 and a3=−12 and a4=−12a2−a1=−12−(−12)=−12+12=0a3−a2=−12−(−12)=−12+12=0a4−a3=−12−(−12)=−12+12=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

a5=a4+d=−12+0=−12a6=a5+d=−12+0=−12a7=a6+d=−12+0=−12

Therefore, the next three terms of given series are −12,−12,−12

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

Answer:

Given series is

1,3,9,27,…

Now, the first term to this series is =1
Now,

a1=1 and a2=3 and a3=9 and a4=27a2−a1=3−1=2a3−a2=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. a,2a,3a,4a,…

Answer:

Given series is

a,2a,3a,4a,…

Now, the first term to this series is = a
Now,

a1=a and a2=2a and a3=3a and a4=4aa2−a1=2a−a=aa3−a2=3a−2a=aa4−a3=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

a5=a4+d=4a+a=5aa6=a5+d=5a+a=6aa7=a6+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. a,a2,a3,a4,…
Answer:

Given series is

a,a2,a3,a4,…

Now, the first term to this series is = a
Now,

a1=a and a2=a2 and a3=a3 and a4=a4a2−a1=a2−a=a(a−1)a3−a2=a3−a2=a2(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. 2,8,18,32,…

Answer:

Given series is

2,8,18,32,…

We can rewrite it as

2,22,32,42,…

Now,
first term to this series is = a
Now,

a1=2 and a2=22 and a3=32 and a4=42a2−a1=22−2=2a3−a2=32−22=2a4−a3=42−32=2

We can clearly see that difference between terms are equal and equal to 2
Hence, given series is in AP
Now, the next three terms are

a5=a4+d=42+2=52a6=a5+d=52+2=62a7=a6+d=62+2=72

Therefore, next three terms of given series are 52,62,72
That is the next three terms are 50,72,98

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

Answer:

Given series is

3,6,9,12,…

Now, the first term to this series is =3
Now,

a1=3 and a2=6 and a3=9 and a4=12a2−a1=6−3=3(2−1)a3−a2=3−3=3(3−1)

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. 12,32,52,72,…

Answer:

Given series is

12,32,52,72,…

we can rewrite it as

1,9,25,49,…

Now, the first term to this series is =1
Now,

a1=1 and a2=9 and a3=25 and a4=49a2−a1=9−1=8a3−a2=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. 12,52,72,73,…

Answer:

Given series is

12,52,72,73,…

we can rewrite it as

1,25,49,73…

Now,
the first term to this series is =1
Now,

a1=1 and a2=25 and a3=49 and a4=73a2−a1=25−1=24a3−a2=49−25=24a4−a3=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

a5=a4+d=73+24=97a6=a5+d=97+24=121a7=a6+d=121+24=145

Therefore, the next three terms of given series are 97,121,145

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

Answer:

(i)

It is given that

a=7,d=3,n=8

Now, we know that

an=a+(n−1)da8=7+(8−1)3=7+7×3=7+21=28

Therefore,

a8=28

(ii) It is given that

a=−18,n=10,a10=0

Now, we know that

an=a+(n−1)da10=−18+(10−1)d0+18=9dd=189=2

(iii) It is given that

d=−3,n=18,a18=−5

Now, we know that

an=a+(n−1)da18=a+(18−1)(−3)−5=a+17×(−3)a=51−5=46

Therefore,

a=46

(iv) It is given that

a=−18.9,d=2.5,an=3.6

Now, we know that

an=a+(n−1)dan=−18.9+(n−1)2.53.6+18.9=2.5n−2.5n=22.5+2.52.5=252.5=10

Therefore,

n=10

(v) It is given that

a=3.5,d=0,n=105

Now, we know that

an=a+(n−1)da105=3.5+(105−1)0a105=3.5

Therefore,

a105=3.5

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

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

Answer:

10,7,4,…

Here, a=10
and

d=7−10=−3

Now, we know that

an=a+(n−1)d

It is given that n=30
Therefore,

a30=10+(30−1)(−3)a30=10+(29)(−3)a30=10−87=−77

Therefore, 30 th term of the AP: 10,7,4,…, is −77

Hence, Correct answer is (C)

Q2 (ii) Choose the correct choice in the following and justify : 11 th term of the AP: −3,−12,2,…, is

(A) 28 (B) 22 (C) -38 (D) −4812

Answer:

Given series is

−3,−12,2,…

Here, a=−3
and

d=−12−(−3)=−12+3=−1+62=52

Now, we know that

an=a+(n−1)d

It is given that n=11
Therefore,

a11=−3+(11−1)(52)a11=−3+(10)(52)a11=−3+5×5=−3+25=22

Therefore, 11 th term of the AP: −3,−12,2,…, is 22

Hence, the Correct answer is (B)

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

Given AP is

3,8,13,18,…

Let suppose that nth term of AP is 78
Here, a=3
And

d=a2−a1=8−3=5

Now, we know that that

an=a+(n−1)d⇒78=3+(n−1)5⇒78−3=5n−5⇒n=75+55=805=16

Therefore, value of 16th term of given AP is 78

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

Answer:

Given AP series is

7,13,19,…,205

Let's suppose there are n terms in given AP
Then,

a=7,an=205

And

d=a2−a1=13−7=6

Now, we know that

an=a+(n−1)d⇒205=7+(n−1)6⇒205−7=6n−6⇒n=198+66=2046=34

Therefore, there are 34 terms in given AP

Q5 (ii) Find the number of terms in each of the following APs : 18,1512,13,…,−47

Answer:

18,1512,13,…,−47
suppose there are n terms in given AP
Then,

a=18,an=−47

And

d=a2−a1=312−18=31−362=−52

Now, we know that

an=a+(n−1)d⇒−47=18+(n−1)(−52)⇒−47−18=−5n2+52⇒−5n2=−65−52⇒−5n2=−1352⇒n=27

Therefore, there are 27 terms in given AP

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

Answer:

Given AP series is

11,8,5,2…

Here, a=11
And

d=a2−a1=8−11=−3

Now, suppose - 150 is nth term of the given AP
Now, we know that

an=a+(n−1)d⇒−150=11+(n−1)(−3)⇒−150−11=−3n+3⇒=n=161+33=1643=54.66

Value of n is not an integer
Therefore, -150 is not a term of AP 11,8,5,2…

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

Answer:

It is given that
11 th term of an AP is 38 and the 16 th term is 73
Now,

Now,

a11=38=a+10d

And

a16=73=a+15d

On solving equation (i) and (ii) we will get

a=−32 and d=7

Now,

a31=a+30d=−32+30×7=−32+210=178

Therefore, 31st terms of given AP is 178

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

It is given that

AP consists of 50 terms of which 3 rd term is 12 and the last term is 106
Now,

a3=12=a+2d

And

a50=106=a+49d

On solving equation (i) and (ii) we will get

a=8 and d=2

Now,

a29=a+28d=8+28×2=8+56=64

Therefore, 29th term of given AP is 64

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

It is given that

3 rd and the 9 th terms of an AP are 4 and -8 respectively
Now,

a3=4=a+2d

And

a9=−8=a+8d

On solving equation (i) and (ii) we will get

a=8 and d=−2

Now,
Let nth term of given AP is 0
Then,

an=a+(n−1)d0=8+(n−1)(−2)2n=8+2=10n=102=5

Therefore, 5th term of given AP is 0

Q10 The 17 th term of an AP exceeds its 10 th term by 77 . Find the common difference.
Answer:

It is given that

17 th term of an AP exceeds its 10 th term by 7
i.e.

a17=a10+7⇒a+16d=a+9d+7⇒a+16d−a−9d=7⇒7d=7⇒d=1

Therefore, the common difference of AP is 1

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

Given AP is

3,15,27,39,…

Here, a=3
And

d=a2−a1=15−3=12

Now, let's suppose nth term of given AP is 132 more than its 54 th term
Then,

an=a54+132⇒a+(n−1)d=a+53d+132⇒3+(n−1)12=3+53×12+132⇒12n=3+636+132+12⇒12n=636+132+12⇒n=78012=65

Therefore, 65th term of given AP is 132 more than its 54 th term

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

It is given that
Two APs have the same common difference and difference between their 100 th terms is 100
i.e.

a100−a100′=100

Let common difference of both the AP's is d

⇒a+99d−a′−99d=100⇒a−a′=100

Now, difference between 1000th term is

a1000−a1000′⇒a+999d−a′−999d⇒a−a′⇒100

Therefore, difference between 1000th term is 100

Q 13 How many three-digit numbers are divisible by 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 an=994

Let there are n three digit numbers divisible by 7
Now, we know that

an=a+(n−1)d⇒994=105+(n−1)7⇒7n=896⇒n=8967=128

Therefore, there are 128 three-digit numbers divisible by 7

Q14 How many multiples of 4 lie between 10 and 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 an=248

Let there are n numbers divisible by 4
Now, we know that

an=a+(n−1)d⇒248=12+(n−1)4⇒4n=240⇒n=2404=60

Therefore, there are 60 numbers between 10 to 250 that are divisible by 4

Q15 For what value of π−i, are the π th terms of two APs: 63,65,67,… and 3,10,17,… equal?

Answer:

Given two AP's are

63,65,67,… and 3,10,17,…

Let first term and the common difference of two AP's are a,a′ and d,d′

a=63,d=a2−a1=65−63=2

And

a′=3,d′=a2′−a1′=10−3=7

Now,
Let nth term of both the AP's are equal

an=an′⇒a+(n−1)d=a′+(n−1)d′⇒63+(n−1)2=3+(n−1)7⇒5n=65⇒n=655=13

Therefore, the 13th term of both the AP's are equal

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

It is given that
3rd term of AP is 16 and the 7 th term exceeds the 5 th term by 12
i.e.

a3=a+2d=16

And

a7=a5+12a+6d=a+4d+122d=12d=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 20 th term from the last term of the AP:3,8,13,…,253

Answer:

Given AP is

3,8,13,…,253

Here, a=3 and an=253
And

d=a2−a1=8−3=5

Let suppose there are n terms in the AP
Now, we know that

an=a+(n−1)d253=3+(n−1)55n=255n=51

So, there are 51 terms in the given AP and 20th term from the last will be 32th term from the starting Therefore,

a32=a+31da32=3+31×5=3+155=158

Therefore, 20th term from the of given AP is 158

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

It is given that
sum of the 4 th and 8 th terms of an AP is 24 and the sum of the 6 th and 10 th terms is 44
i.e.

a4+a8=24⇒a+3d+a+7d=24⇒2a+10d=24⇒a+5d=12

And

a6+a10=44⇒a+5d+a+9d=44⇒2a+14d=44⇒a+7d=22

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

an=a+(n−1)d⇒7000=5000+(n−1)200⇒2000=200n−200⇒200n=2200⇒n=11

Therefore, after 11 years 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 1.75 . If in the n th week, her weekly savings become Rs 20.75 , find 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 1.75
Therefore, a=5 and d=1.75
after nth week, her weekly savings become Rs 20.75
Now, we know that

an=a+(n−1)d⇒20.75=5+(n−1)1.75⇒15.75=1.75n−1.75⇒1.75n=17.5⇒n=10

Therefore, after 10 weeks her saving will become Rs 20.75

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

Given AP is
2,7,12,…, to 10 terms
Here, a=2 and n=10
And

d=a2−a1=7−2=5

Now, we know that

S=n2{2a+(n−1)d}⇒S=102{2×2+(10−1)5}⇒S=5{4+45}⇒S=5{49}⇒S=245

Therefore, the sum of AP 2,7,12,…, to 10 terms is 245

Q1 (ii) Find the sum of the following APs: −37,−33,−29,…, to 12 terms.

Answer:

Given AP is
−37,−33,−29,…, to 12 terms.
Here, a=−37 and n=12
And

d=a2−a1=−33−(−37)=4

Now, we know that

S=n2{2a+(n−1)d}⇒S=122{2×(−37)+(12−1)4}⇒S=6{−74+44}⇒S=5{−30}⇒S=−180

Therefore, the sum of AP−37,−33,−29,…, to 12 terms. is −180

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

Answer:

Given AP is
0.6,1.7,2.8,…, to 100 terms..
Here, a=0.6 and n=100
And

d=a2−a1=1.7−0.6=1.1

Now, we know that

S=n2{2a+(n−1)d}⇒S=1002{2×(0.6)+(100−1)(1.1)}⇒S=50{1.2+108.9}⇒S=50{110.1}⇒S=5505

Therefore, the sum of AP0.6,1.7,2.8,…, to 100 terms. is 5505

Q1 (iv) Find the sum of the following APs: 115,112,110,…, to 11 terms.

Answer:

Given AP is
115,112,110,…, to 11 terms.
Here, a=115 and n=11
And

d=a2−a1=112−115=5−460=160

Now, we know that

S=n2{2a+(n−1)d}⇒S=112{2×115+(11−1)(160)}⇒S=112{215+16}⇒S=112{930}⇒S=9960=3320

Therefore, the sum of AP 115,112,110,…, to 11 terms. is 3320

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

Answer:

Given AP is

121,117,113,…

Here a=121 and d=−4
Let suppose nth term of the AP is first negative term
Then,

an=a+(n−1)d

If n nth term is negative then an<0

⇒121+(n−1)(−4)<0⇒125<4n⇒n>1254=31.25

Therefore, first negative term must be 32nd term

Q2 The sum of the third and the seventh terms of an AP is 6 and their product is 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 8

a3=a+2da7=a+6d

Now,

a3+a7=a+2d+a+6d=6⇒2a+8d=6⇒a+4d=3⇒a=3−4d

And

a3⋅a7=(a+2d)⋅(a+6d)=a2+8ad+12d2=8−(ii)

put value from equation (i) in (ii) we will get

⇒(3−4d)2+8(3−4d)d+12d2=8⇒9+16d2−24d+24d−32d2+12d2=8⇒4d2=1⇒d=±12

Now,

case (i)d=12

a=3−4×12=1

Then,

S16=162{2×1+(16−1)12}

S16=76

case (ii) d=−12a=3−4×(−12)=5
Then,

S16=162{2×1+(16−1)(−12)}S16=20

Q3 A ladder has rungs [25]cm apart. (see Fig_ [5.7 ). The rungs decrease uniformly in length from 450 cm at the bottom to [55 cm at the top. If the top and the bottom rungs are 22] m apart, what is the length of the wood required for the rungs? [ Hint: Number of gs=25025+1]

Answer:

It is given that
The total distance between the top and bottom rung =212 m=250 cm
Distance between any two rungs =25 cm
Total number of rungs =25025+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,a11=45 and n=11
Now, we know that

a11=a+10d

⇒45=25+10d⇒d=2

Now, total length of the wood required for the rungs is equal to

S11=112{2×25+(11−1)2}S11=112{50+20}S11=112×70S11=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 [1] to 49 . Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it Find this value x.[ Hint :Sx−1=S49−Sx]

Answer:

It is given that the sum of the numbers of the houses preceding the house numbered 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

Sn=n2{2a+(n−1)d}
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.
Sn−1=S49−Sn
n−12{2a+((n−1)−1)d}=492{2a+(49−1)d}−n2{2a+(n−1)d}
n−12{2+(n−2)}=492{2+48}−n2{2+(n−1)}
n−12{n}=492{50}−n2{n+1}

n22−n2=1225−n22−n2

n2=1225n=±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

Q5A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 14m and a tread of 12m (see Fig , 5.8 ). Calculate the total volume of concrete required to build the terrace. Hint: Volume of concrete required to build the first step =14×12×50 m3,

Answer:

It is given that

football ground comprises of 15 steps each of which is 50 m long and Each step has a rise of 14m and a tread of 12m
Now,
The volume required to make the first step =14×12×50=6.25 m3

Similarly,
The volume required to make 2nd step =(14+14)×12×50=12×12×50=12.5 m3
And
The volume required to make 3 rd step =(14+14+14)×12×50=34×12×50=18.75 m3

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

S15=152{2×6.25+(15−1)6.25}S15=152{12.5+87.5}S15=152×100S15=15×50=750

Therefore, the total volume of concrete required to build the terrace of 15 such steps is 750 m3