## Arithmetic Progression Practice Questions: how to find missing terms of AP?

**CBSE NCERT Solutions Chapter 5 Arithmetic Progressions Exercise 5.2 Question 3**

**In the following AP's find the missing terms:**

**(i) 2, __ , 26**

**(ii) __, 13, __, 3**

**(iii) 5, __, __, **

**(iv) -4. __, __, __, __, 6**

**(v) __, 38, __, __, __, -22**

**Solution (i)**

**2, __ , 26**

We know that difference between consecutive terms is constant in any Arithmetic progression. **(1)**

Let the missing term be .

Using property **(1)**, we can say that

Therefore, missing term is .

**Solution (ii)**

** __, 13, __, 3**

Let missing terms be and .

The sequence becomes:

We know that difference between consecutive terms is constant in any Arithmetic progression. **(1)**

Using property **(1)**, we can say that

Also, using property **(1), **we can say that

But, we have , Putting value of in the above equation, we get

**Therefore, missing terms are 18 and 8.**

**Solution (iii)**

We can also use the following method to solve similar problems. It is also easy. The first two questions can also be solved by the same method.

**5, __, __, **

Here, first term =

And, 4th term =

Using formula , to find nth term of arithmetic progression, we get

Therefore, we get common difference = d =

Second term = a+ d =

Third term = second term + d =

**Therefore, missing terms are and **

**Solution (iv)**

** -4. __, __, __, __, 6**

Here, First term = a = -4

6th term = = 6

Using formula , to find nth term of arithmetic progression, we get

Therefore, common difference = d = 2

Second term = first term + d = a+ d = -4 + 2 = -2

Third term = second term + d = -2 + 2 = 0

Fourth term = third term + d = 0 + 2 = 2

Fifth term = fourth term + d = 2 + 2 = 4

**Therefore, missing terms are -2, 0, 2 and 4.**

**Solution (v) **

**__, 38, __, __, __, -22**

In this problem, we are given 2nd and 6th term.

Using formula , to find nth term of arithmetic progression, we get

And

And

These are equations in two variables, we can solve them using any method. Lets solve them using substitution method.

Using equation (), we can say that . Putting value of in equation (), we get

Using this value of and putting this in equation , we get

Therefore, we get and

First term = a = 53

Third term = second term + d = 38 - 15 = 23

Fourth term = third term + d = 23 - 15 = 8

Fiftht term = fourth term + d = 8 - 15 = -7

**Therefore, missing terms are 53, 23, 8 and -7.**