Question

find the sum of the first 14 terms of an A.P 3, 15, 27, 39,.........​

Answer 1

Given:-

AP: 3,15,27,39,.....

To find:-

Sum of the first 14 terms of the given AP

Formula used:-

$S_n = \frac{n}{2}[(2a_1+(n-1)d]$

Solution:-

We have,

$a_1 = 3$ , d = 15 - 3 = 12 and, n = 14

By using the formula,

$S_{14} = \frac{14}{2}$ [2 × 3 +13 ×15]

     = 7(6 + 195)

     = 7 × 201

     = 1407

Hence, the sum of first 14 terms of the AP = 1407

More to know:-

$n^{th}$ term of an AP, $a_n$ = $a_1$ + (n-1)d

d = $a_2-a_1$

Answer 2

Answer:

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