Question

sum of first 10 terms of an ap 3, 15, 27, 39​

Answer 1

Answer:

n=10

given a= 3

d = 15-3= 12

sum of n terms of an ap =n/2[2a+(n-1)d]

Sn=10/2[2*3 +(10-1)12]

S=5[6+108]

S=5(114)

S=570. answer

Step-by-step explanation:

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Answer 2

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Given :-

A.P is 3,15,27,39

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To find :-

Sum of first 10 terms

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Solution

First term (a) = 3

Common Difference (d) = 27-15 => 12

We have formula :-

$\huge{\boxed{\sf{S_{n} \: = \: \frac{n}{2}[2a \: + \: (n-1)d]}}}$

_______[Put Values]

⇒S10 = 10/2 × [2(3) + (10-1)12]

⇒S10 = 5 × [6 + 108]

⇒S10 = 5 × (114)

⇒ S10 = 570

∴ Sum of 10 terms are 570