The sum of the first 12 terms of an AP. Whose nth term is given by an = 3n + 4 is:
1. 262
2. 272
3. 282
4. 292
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Correct option is 3. 282
an = 3n + 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