Answer :

Formula used.

S_{n} = [2a + (n–1)d]

d = a_{n + 1}–a_{n}

In the following A.P

a = 5,

d = a_{2}–a_{1}

= 7–5 = 2;

n = 30

S_{n} = [2a + (n–1)d]

= [2×5 + (30–1)2]

= 15[10 + 29×2]

= 15[10 + 58]

= 1020

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