Question

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

Answer 1

Answer:

1134

Step-by-step explanation:

given AP is 3, 15, 27, 39 ........

a = 3

d = a2 - a1 = 15 - 3 = 12

the sum of 14 terms

n = 14

Sn = n/2 (2a + (n-1)d)

$s14= \frac{14}{2} (2(3) + (14 - 1)(12)) \\ = 7(6 + 13(12)) \\ = 7 (6 + 156) \\ = 7 \times 162 \\ = 1134$

the sum of 14 terms of the AP is 1134

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