Question

The sum of the first 12 terms of an AP. Whose nth term is given by a_n = 3_n + 4 is: 

1. 262 

2. 272 

3. 282 

4. 292

Answer 1

Correct option is 3. 282

a_n = 3_n + 4

Put n = 1, 1, 3 [series will be 7, 10, 13, …] 

a = 7, d = 2, n = 12

\(s_{12}=\frac n2[2a+(n-1)d]\)

\(= \frac{12}2[2(7)+(12-3)3]\)

= 6(14 + 39)

= 282