Question

How many terms are present in the sequence of A.P. 6, 11, 16, 21, ......... whose sum is 969?

Answer 1

Given sequence is 6, 11, 16, 21, .....

Let a be the first term and d be the common difference.

Then, a = 6, d = 11 – 6 = 5

We know, `S_n = n/2 [2a + (n - 1)d]`

⇒ 969 = `n/2 [2(6) + (n - 1) (5)]`

⇒ 969 = `n/2 [12 + 5n - 5]`

⇒ 969 = `n/2 [7 + 5n]`

⇒ 1938 = 7n + 5n²

⇒ 5n² + 7n – 1938 = 0

⇒ 5n² + (102 – 95)n – 1938 = 0

⇒ 5n² + 102n – 95n – 1938 = 0

⇒ n(5n + 102) – 19(5n + 102) = 0

⇒ (5n + 102) (n – 19) = 0

n = 19, – `102/5`

The negative value is rejected because the number of words cannot be negative.

As a result, the A.P. contains 19 terms.