# 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

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

It is given that
Fare for   = Rs. 15
And after that   for each additional
Now,
Fare for   = Fare of first km + Additional fare for 1 km
=  Rs. 15 + 8 = Rs 23

Fare for  = Fare of first km + Fare of additional second km + Fare of additional third km
= Rs. 23 + 8= Rs 31

Fare of n km =
( 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

It is given that
vacum pump removes     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

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

Now,

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

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

It is given that
Cost of digging of 1st meter = Rs 150
and
rises by    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

Amount in the beginning = Rs. 10000

Interest at the end of 1st year at the rate of
is  of 10000 =

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

= 10000 + 800

= 10800

Now,

Interest at the end of 2nd year at rate of
is   of 10800 =

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

It is given that

Now,

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:Answer:

It is given that

Now,

Therefore, the first four terms of the given series are -2,-2,-2,-2

It is given that

Now,

Therefore, the first four terms of the given series are 4,1,-2,-5

It is given that

Now,

Therefore, the first four terms of the given series are

It is given that

Now,

Therefore, the first four terms of the given series are  -1.25,-1.50,-1.75,-2

Given AP series is

Now, first term of this AP series is 3

Therefore,

First-term of AP series (a) = 3

Now,

And common difference (d)  =

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

Given AP series is

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

Therefore,

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

Now,

And common difference (d)  =

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

Given AP series is

Now, the first term of this AP series is

Therefore,

The first term of AP series (a) =

Now,

And common difference (d)  =

Therefore, the first term and the common difference is  and   respectively

Given AP series is

Now, the first term of this AP series is 0.6

Therefore,

First-term of AP series (a) = 0.6

Now,

And common difference (d)  =

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

Given series is

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

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

Given series is

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

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

Therefore, next three terms of given series are

Given series is

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

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

Therefore, next three terms of given series are  -9.2,-11.2,-13.2

Given series is

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

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, next three terms of given series are  6,10,14

Given series is

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

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

Therefore, next three terms of given series are

Given series is

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

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

Given series is

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

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

Given series is

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

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

Therefore, the next three terms of given series are

Given series is

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

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

Given series is

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

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

Therefore, next three terms of given series are  5a,6a,7a

Given series is

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

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

Given series is

We can rewrite it as

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

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

Therefore, next three terms of given series are

That is the next three terms are

Given series is

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

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

Given series is

we can rewrite it as

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

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

Given series is

we can rewrite it as

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

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

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

 a d n (i) (ii) (iii) (iv) (v) 7 3 0 8 10 18 105 0

(i)
It is given that

Now,  we know that

Therefore,

(ii) It is given that

Now,  we know that

(iii) It is given that

Now,  we know that

Therefore,

(iv) It is given that

Now,  we know that

Therefore,

(v) It is given that

Now,  we know that

Therefore,

(A)              (B)              (C)              (D)

Given series is

Here,
and

Now, we know that

It is given that
Therefore,

Therefore, th term of the AP:     is -77

(A)               (B)               (C)               (D)

Given series is

Here,
and

Now, we know that

It is given that
Therefore,

Therefore, 11th term of the AP:     is 22
Hence, the Correct answer is (B)

Given AP series is

Here,
Now, we know that

Now,

Therefore, the missing term is 14

Given AP series is

Here,
Now,

Now, we know that

Now,

And

Therefore, missing terms are  18 and 8
AP series is 18,13,8,3

Given AP series is

Here,
Now, we know that

Now,

And

Therefore, missing terms are   and 8
AP series is

## Q3 (iv) In the following APs, find the missing terms in the boxes :

Given AP series is

Here,
Now, we know that

Now,

And

And

And

Therefore, missing terms are   -2,0,2,4
AP series is -4,-2,0,2,4,6

Given AP series is

Here,
Now,

Now, we know that

Now,

And

And

And

Therefore, missing terms are   53,23,8,-7
AP series is 53,38,23,8,-7,-22

Given AP is

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

Now, we know that that

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

Given AP series is

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

And

Now, we know that

Therefore, there are 34 terms in given AP

Given AP series is

suppose there are n terms in given AP
Then,

And

Now, we know that

Therefore, there are 27 terms in given AP

Given AP series is

Here,
And

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

Value of n is not an integer
Therefore, -150 is not a term of AP

It is given that
th term of an AP is   and the th term is
Now,

And

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

Now,

Therefore, 31st terms of given AP is 178

It is given that
AP consists of terms of  which rd term is and the last term is
Now,

And

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

Now,

Therefore, 29th term of given AP is 64

It is given that
rd and the th terms of an AP are   and    respectively
Now,

And

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

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

Therefore, 5th term of given AP is 0

It is given that
th term of an AP exceeds its th term by
i.e.

Therefore, the common difference of AP is 1

Given  AP is

Here,
And

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

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

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

Let common difference of both the AP's is d

Now, difference between 1000th term is

Therefore, difference between 1000th term is 100

We know that the first three digit number divisible by 7 is 105 and last three-digit number divisible by 7 is 994
Therefore,

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

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

We know that the first number divisible by 4 between 10 to 250 is 12 and last  number divisible by 4 is 248
Therefore,

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

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

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

And

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

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

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

And

Put the value of d in equation (i) we will get

Now, AP with first term = 4 and common difference = 6 is
4,10,16,22,.....

Given AP is

Here,
And

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

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

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

It is given that
sum of the  th and  th terms of an AP is   and the sum of the th and th terms is
i.e.

And

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

Therefore,first three of AP with a = -13 and d = 5 is
-13,-8,-3

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,
Let's suppose after n years his salary will be Rs 7000
Now, we know that

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

It is given that
Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by  Rs
Therefore,
after  th week, her weekly savings become Rs
Now, we know that

Therefore, after 10 weeks her saving will become Rs 20.75

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

Given AP is
to  terms
Here,
And

Now, we know that

Therefore, the sum of AP   to  terms is 245

Given AP is
to  terms.
Here,
And

Now, we know that

Therefore, the sum of AP   to  terms. is -180

Given AP is
to   terms..
Here,
And

Now, we know that

Therefore, the sum of AP  to   terms. is 5505

Given AP is
to     terms.
Here,
And

Now, we know that

Therefore, the sum of AP    to     terms. is

Given AP is

We first need to find the number of terms
Here,
And

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

Now, we know that

Therefore, the sum of AP    is

Q2 (ii) Find the sums given below :

Given AP is

We first need to find the number of terms
Here,
And

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

Now, we know that

Therefore, the sum of AP    is 286

Q2 (iii) Find the sums given below :

Given AP is

We first need to find the number of terms
Here,
And

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

Now, we know that

Therefore, the sum of AP    is -8930

It is given that

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

Now, we know that

Therefore, the sum of the given AP  is 440

Q3 (ii) In an AP: given , find and .

It is given that

Now, we know that

Therefore, the sum of given AP  is 273

Q3 (iii) In an AP: given   find  and .

It is given that

Now, we know that

Therefore, the sum of given AP  is 246

Q3 (iv) In an AP: given  find  and

It is given that

Now, we know that

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

Now,

Therefore, the value of d and 10th terms is -1 and 8 respectively

Q3 (v) In an AP: given , find and .

It is given that

Now, we know that

Now,

Q3 (vi) In an AP: given   find and .

It is given that

Now, we know that

n can not be negative so the only the value of n is 5
Now,

Therefore, value of n and nth term is 5 and 34 respectively

Q3 (vii) In an AP: given   find and .

It is given that

Now, we know that

Now, we know that

Now, put this value in (i) we will get

Therefore, value of n and d are  respectively

Q3 (viii) In an AP: given   find  and .

It is given that

Now, we know that

Now, we know that

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    find .

It is given that

Now, we know that

Therefore, the value of d is 6

It is given that

Now, we know that

Now, we know that

Therefore, the value of a is 4

Given AP is

Here,
And
Now , we know that

Value of  n  can not be negative so the only the value of n is 12
Therefore, the sum of 12 terms of AP   must be taken to give a sum of .

It is given that

Now, we know that

Now, we know that

Now, put this value in (i) we will get

Therefore, value of n and d are  respectively

It is given that

Now, we know that

Now, we know that

Therefore, there are 38 terms  and their sun is 6973

It is given that

Now, we know that

Now, we know that

Therefore, there are 22 terms  and their sum is 1661

It is given that

And
Now,

Now, we know that

Therefore, there are 51 terms  and their sum is 5610

It is given that

Now, we know that

Similarly,

On solving equation (i) and (ii) we will get
a = 1 and d = 2
Now, the sum of first n terms is

Therefore, the sum of n terms  is

It is given that

We will check values of  for different values of n

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

Therefore, the sum of 15 terms  is  525

It is given that

We will check values of  for different values of n

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

Therefore, the sum of 15 terms is -465

It is given that
the sum of the first terms of an AP is
Now,

Now, first term is

Therefore, first term is 3
Similarly,

Therefore, sum of first two terms is 4
Now, we know that

Now,

Similarly,

Positive integers divisible by 6 are
6,12,18,...
This is an AP with

Now, we know that

Therefore,  sum of the first positive integers divisible by   is 4920

First 15 multiples of 8 are
8,16,24,...
This is an AP with

Now, we know that

Therefore,  sum of the first 15 multiple of 8 is 960

The odd number between 0 and 50 are
1,3,5,...49
This is an AP with

There are total 25 odd number between 0 and 50
Now, we know that

Therefore,  sum of the odd numbers between and   625

It is given that
Penalty for delay of completion beyond a certain date is Rs   for the first day, Rs  for the second day, Rs  for the third day and  penalty for each succeeding day being Rs  more than for the preceding day
We can clearly see that
200,250,300,..... is an AP  with

Now, the penalty for 30 days is given by the expression

Therefore, the penalty for 30 days is 27750

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
Let a be the prize money given to the 1st student
Then,

Therefore, the prize given to the first student is Rs 160
Now,
Let  is the prize money given to the next 6 students
then,

Therefore, prize money given to 1 to 7 student is 160,140,120,100,80,60.40

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

Therefore,  number of trees planted by 12 classes is 234 [Hint : Length of successive semicircles is    with centres at   respectively.]

From the above-given figure

Circumference of 1st semicircle

Similarly,

Circumference of 2nd semicircle

Circumference of 3rd semicircle

It is clear that this is an AP with

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

Therefore, sum of length of 13 such semicircles is 143 cm 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
Let suppose 200 logs are arranged in 'n' rows,
Then,

Now,
case (i) n = 25

But number of rows can not be in negative numbers
Therefore, we will reject the value n = 25

case (ii) n = 16

Therefore, the number of rows in which 200 logs are arranged is equal to 5 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    ]

Distance travelled by the competitor in picking and dropping 1st potato

Distance travelled by the competitor in picking and dropping 2nd potato

Distance travelled by the competitor in picking and dropping 3rd potato

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

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

Given AP is

Here
Let suppose nth term of the AP is first negative term
Then,

If nth term is negative then

Therefore, first negative term must be 32nd term

It is given that sum of third and seventh terms of an AP are and their product is

Now,

And

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

Now,
case (i)

Then,

case (ii)

Then, It is given that
The total distance between the top and bottom rung
Distance between any two rungs = 25 cm
Total number of rungs =
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
Now, we know that

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

Therefore, the total  length of the wood required for the rungs is equal to  385 cm

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

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.

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 It is given that
football ground comprises of   steps each of which is m long and Each step has a rise of      and a tread of
Now,
The volume required to make the first step =

Similarly,

The volume required to make 2nd step =
And
The volume required to make 3rd step =

And so on
We can clearly see that this is an AP with
Now, the total volume of concrete required to build the terrace of 15 such step is

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

## 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

## 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!