Question

How many terms of the AP 3, 7, 11, 15, ... will make the sum 406?

Options

• 10

• 12

• 14

• 20

Answer 1

14

Explanation:

Let S_n = 406.

Then, `n/2 [2 xx 3 + (n - 1) xx 4] = 406`

⇒ `n/2 (6 + 4n - 4) = 406`

⇒ `n/2 (4n + 2) = 406`

⇒ n(2n + 1) = 406

⇒ 2n² + n – 406 = 0

∴ `n = (-1 +- sqrt(1 + 3248))/4`

= `(-1 + sqrt(3249))/4`   ...[Neglecting negative value]

⇒ `n = (-1 + 57)/4`

= `56/4`

= 14