The first exercise of chapter 5, class 10, Arithmetic progression, is the introduction to the same which discusses patterns we come face-to-face in our everyday life where the succeeding term results from adding a fixed number to the preceding terms. The exercise also talks about the meaning of Arithmetic Progressions and other related terms, along with examples. The general form of AP is a, a + d, a + 2d, a + 3d,…

Example- 

The taxi fare after each km when it is the fare is Rs 15 for the first km and Rs 8 for each additional km.

Solution:

We can write the given condition as;

Taxi fare for 1 km = 15

Taxi fare for first 2 km = 15+8 = 23

Taxi fare for first 3 km = 23+8 = 31

Taxi fare for first 4 km = 31+8 = 39

And so on……

Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.

 

The second exercise of class 10, Arithmetic progression is finding the nth Term Of An AP. Here, different concepts are explained suitable for finding the nth term of an AP and related examples.

Example- 

Find the number of terms in each of the following A.P.

(i) 7, 13, 19, …, 205

Solutions:

(ii) Given, 7, 13, 19, …, 205 is the A.P

Therefore

First term, a = 7

Common difference, d = a2 − a1 = 13 − 7 = 6

Let there are n terms in this A.P.

an = 205

As we know, for an A.P.,

an = a + (n − 1) d

Therefore, 205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

Therefore, this given series has 34 terms in it.

 

The third exercise of chapter 5 class 10 Arithmetic progression talks about the various tricks or methods to identify the nth term of an AP. There are different concepts explained through several examples for finding the nth term of an AP.

Example- 

Find the sum of 0.6, 1.7, 2.8 ,…….., to 100 terms

Given, 0.6, 1.7, 2.8 ,…, to 100 terms

For this A.P.,

first term, a = 0.6

Common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1

n = 100

We know that, the formula for sum of nth term in AP series is,

Sn = n/2[2a +(n-1)d]

S12 = 50/2 [1.2+(99)×1.1]

= 50[1.2+108.9]

= 50[110.1]

= 5505

The last exercise of class 10 chapter 5 Arithmetic progression is the summary of the chapter which gives an overview of what all we discussed and explains the entire chapter in an easy-to-understand language. By going through it, one can easily recall all concepts which have been covered in all exercises.

1- Write first four terms of the A.P. when the first term a and the common difference are given as follows:

(i) a = 10, d = 10

(ii) a = -2, d = 0

Solutions:

(i) a = 10, d = 10

Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …

a1 = a = 10

a2 = a1+d = 10+10 = 20

a3 = a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

a5 = a4+d = 40+10 = 50

And so on…

Therefore, the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of this A.P. will be 10, 20, 30, and 40.

2- Which term of the A.P. 3, 8, 13, 18, … is 78?

Solutions:

Given the A.P. series as3, 8, 13, 18, …

First term, a = 3

Common difference, d = a2 − a1 = 8 − 3 = 5

Let the nth term of given A.P. be 78. Now as we know,

an = a+(n−1)d

Therefore,

78 = 3+(n −1)5

75 = (n−1)5

(n−1) = 15

n = 16

Hence, the 16th term of this A.P. is 78.

3- If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

Solution:

Given that,

3rd term, a3 = 4

and 9th term, a9 = −8

We know that,

an = a+(n−1)d

Therefore,

a3 = a+(3−1)d

4 = a+2d ……………………………………… (i)

a9 = a+(9−1)d

−8 = a+8d ………………………………………………… (ii)

On subtracting equation (i) from (ii), we will get here,

−12 = 6d

d = −2

From equation (i), we can write,

4 = a+2(−2)

4 = a−4

a = 8

Let the nth term of this A.P. be zero.

an = a+(n−1)d

0 = 8+(n−1)(−2)

0 = 8−2n+2

2n = 10

n = 5

Hence, the 5th term of this A.P. is 0.

4- In an AP

(i) Given a = 5, d = 3, an = 50, find n and Sn.

(ii) Given a = 7, a13 = 35, find d and S13.

 

Solutions:

(i) Given that, a = 5, d = 3, an = 50

As we know, from the formula of the nth term in an AP,

an = a +(n −1)d,

Therefore, putting the given values, we get,

⇒ 50 = 5+(n -1)×3

⇒ 3(n -1) = 45

⇒ n -1 = 15

⇒ n = 16

Now, sum of n terms,

Sn = n/2 (a +an)

Sn = 16/2 (5 + 50) = 440

(ii) Given that, a = 7, a13 = 35

As we know, from the formula of the nth term in an AP,

an = a+(n−1)d,

Therefore, putting the given values, we get,

⇒ 35 = 7+(13-1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Now, Sn = n/2 (a+an)

S13 = 13/2 (7+35) = 27

5- 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 numbers of the houses following it. Find this value of x. [Hint :Sx – 1 = S49 – Sx]

 

Solution:

Given,

Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First-term, a = 1

The Common difference, d=1

Let us say the number of xth houses can be represented as;

Sum of nth term of AP = n/2[2a+(n-1)d]

Sum of number of houses beyond x house = Sx-1

= (x-1)/2[2(1)+(x-1-1)1]

= (x-1)/2 [2+x-2]

= x(x-1)/2 ………………………………………(i)

By the given condition, we can write,

S49 – Sx = {49/2[2(1)+(49-1)1]}–{x/2[2(1)+(x-1)1]}

= 25(49) – x(x + 1)/2 ………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore,

x(x-1)/2 = 25(49) – x(x+1)/2

x = ±35

As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.

     1. What is Arithmetic Progression Class 10? 

Class 10 Arithmetic profession is one of the most important chapters of the CBSE class 10 maths syllabus as it explains the concept of arithmetic progression and theorems around it.

     2. Is Class 10 Arithmetic Progressions of Class 10 Hard?

Although all exercises in chapter 5 arithmetic progressions are well balanced with moderate-level questions, it still becomes difficult for students to understand them. NCERT Class 10 maths solutions have explained every concept in detail, easily clarifying all doubts without brainstorming. 

     3. What are the uses of Arithmetic Progressions ?"

Class 10 AP or Arithmetic Progression can be applied in real life in well-defined patterns. For example, AP can predict the future or assume something to land on reliable outcomes.