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# If the sum of a certain number of terms of the A.P 25, 22, 19, .... is 116 . Find the last term

**Solution:**

It is given that 25, 22,19, ....., are in A.P.

Let the sum of n terms of an A.P be 116 .

Here, first term a = 25

Common difference d = 22 - 25 = - 3

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

Therefore,

⇒ 116 = n/2 [2(25) + (n - 1) (- 3)]

⇒ 116 = n/2 [50 - 3n + 3]

⇒ 232 = n (53 - 3n) = 53n - 3n^{2}

⇒ 3n^{2} - 53n + 232 = 0

⇒ 3n^{2} - 24n - 29n + 232 = 0

⇒ 3n (n - 8) - 29 (n - 8) = 0

⇒ (n - 8)(3n - 29) = 0

⇒ n = 8, 29/3

However, n cannot be equal to 29/3

Hence, n = 8

Therefore, last term l = a_{n} = a + (n - 1) d

a_{8} = 25 + (8 - 1)(- 3)

= 25 + (7)(- 3)

= 25 - 21

= 4

Thus, the last term of the A.P is 4

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 6

## If the sum of a certain number of terms of the A.P 25, 22, 19, .... is 116 . Find the last term

**Summary:**

The last term of the A.P 25, 22, 19... where the sum is 116 is 4

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