Question

find the sum of first 10 terms of AP 3,15,27,39...

Answer 1

a = 3
d = 15-3 = 12
n = 10
Sum
$= \frac{n(2a + (n - 1)d)}{2} \\ = 5(6 + 108) \\ = 570$

Answer 2

Here is Your answer,

Let a be the first tem of this AP
And d be the common difference

a = 3
n = 10
d = 15-3
= 12

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

S10 = 10/2[2x3+(10-1) 12]

= 5(6+9x12)

= 5(6+108)

= 5x114

= 570

Therefore, the sum of first 10 terms is 570.





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