Question

In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P. must be taken for their sum to be equal to 120?

Options

• 6

• 7

• 8

• 9

Answer 1

8

Explanation;

Here a = 1, d = 4, S_n = 120

S_n = `"n"/2 [2"a" + ("n" - 1)"d"]`

120 = `"n"/2 [2 + ("n" - 1)4]`

= `"n"/2 [2 + 4"n" - 4]`

= `"n"/2 [4"n" - 2]`

= `"n"/2 xx 2 (2"n" - 1)`

120 = 2n² – n

∴ 2n² – n – 120 = 0

⇒ 2n² – 16n + 15n – 120 = 0

2n(n – 8) + 15 (n – 8) = 0

⇒ (n – 8) (2n + 15) = 0

n = 8 or n = `(-15)/2` ...(omitted)

∴ n = 8