Answer :

Formula used.

S_{n} = [2a + (n–1)d]

d = a_{n + 1}–a_{n}

In the sum above A.P follows

Is 7, 12, 17, 22, ……, 102

In the following A.P

⇒ a = 7,

⇒ d = 12–7 = 5

As the last term is 102

a_{n} = a + (n–1)d

102 = 7 + (n–1)5

102 = 7 + 5n–5

5n = 102–2

⇒ n = = 20

S_{n} = [2a + (n–1)d]

= [2×7 + (20–1)5]

= 10[14 + 19×5]

= 10[14 + 95]

= 1090

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