The sum of the first 12 terms of an AP. Whose nth term is given by an�

1.	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
Answer» 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

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

Answer»

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

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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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