# How many terms of the A.P. - 6, - 11/2, - 5, .... are needed to give the sum - 25?

**Solution:**

It is given that

- 6, - 11/2, - 5, .... are in A.P.

Let the sum of n terms of the given A.P be, S_{n} = - 25

Here, first term a = -6

Common difference d = - 11/2 + 6

= (- 11 + 12)/2 = 1/2

We know that, S_{n} = n/2 [2a + (n - 1) d]

Therefore,

⇒ - 25 = n/2 [2(- 6) + (n - 1) (1/2)]

⇒ - 50 = n [- 12 + n/2 - 1/2]

⇒ - 50 = n [n/2 - 25/2]

⇒ - 100 = n[n - 25]

⇒ n^{2} - 25n + 100 = 0

⇒ n^{2} - 5n - 20n + 100 = 0

⇒ n (n - 5) - 20 (n - 5) = 0

⇒ (n - 5)(n - 20) = 0

⇒ n = 20, 5

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 4

## How many terms of the A.P. - 6, - 11/2, - 5, .... are needed to give the sum - 25 ?

**Summary:**

Due to the fact that this is an increasing A.P and starts from the negative side, it is possible to have two solutions which were 20 and 5

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