# Class 10 NCERT Solutions Maths Chapter 5 - Arithmetic Progressions

Find complete NCERT Solutions for CBSE Class 10 Mathematics Chapter 5 Arithmetic Progressions at TopperLearning. Our experts will give you the accurate steps to show you how to apply the concept of equating the general term of two arithmetic progressions to solve a corresponding linear equation.

Practise with our NCERT textbook solutions to understand how to work with constants, variables and arithmetic operations to solve algebraic expressions. To learn the basics of arithmetic progressions, watch our concept videos and explore our Maths resources such as sample papers, past years’ question papers etc.

## Arithmetic Progressions Exercise Ex. 5.1

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## Arithmetic Progressions Exercise Ex. 5.2

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We have to find the 30^{th} term and 11^{th} term in I and II respectively.

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## Arithmetic Progressions Exercise Ex. 5.3

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(i)** **2, 7, 12 ,…, to 10 terms

For this AP,

*a* = 2

*d* = *a*_{2} – *a*_{1} = 7 – 2 = 5

*n* = 10

We know that,

(ii)** **–37, –33, –29 ,…, to 12 terms

For this AP,

*a* = –37

*d* = *a*_{2} – *a*_{1} = (–33) – (–37)

= – 33 + 37 = 4

*n* = 12

We know that,

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms

For this AP,

*a* = 0.6

*d* = *a*_{2} – *a*_{1} = 1.7 – 0.6 = 1.1

*n* = 100

We know that,

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(i)

(ii)

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In a potato race, a bucket is placed at the starting point, which is 5m from the first potato and other potatoes are placed 3m apart in a straight line. There are ten potatoes in the line.

So, we get the series as

5, 8, 11, 14, ........

Here, a= 5 and d = 8 - 5 = 11 - 8= 3

the difference between the two consecutive terms are same.

So, this is an arithmetic Progression.

According to the condition, we get the series as

5 + 5 , 8 + 8, 11 + 11, .....

10, 16, 22, ..............

Here a = 10 and d = 16 - 10 = 6

the difference between the two consecutive terms are same.

So, this is an arithmetic Progression.

The total distance the competitor has to run is given by,

Therefore the total distance the competitor has to run is 370 m.

## Arithmetic Progressions Exercise Ex. 5.4

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The number of houses was 1, 2, 3, ..... 49.

It can be observed that the number of houses are in an AP having a as 1 and d also as 1.

Let us assume that the number of x^{th} house was like this.

We know that,

Sum of number of houses preceding x^{th} house = S_{x-1}

Sum of number of houses preceding x^{th} house = S_{49 }- S_{x}

However, the house numbers are positive integers.

The value of x will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding

the house numbered 35 is equal to the sum of the numbers of the houses following it.