Answer :

We know, nth term of an AP is

a_{n} = a + (n – 1)d

where ‘a’ and ‘d’ are first term and common difference of AP respectively

Given, third term of an A.P. is 7

a_{3} = 7

⇒ a + 2d = 7 (i)

the seventh term exceeds three times the third term by 2

⇒ a_{7} = 3a_{3} + 2

⇒ a + 6d = 3(7) + 2

⇒ 7 – 2d +6d = 21 + 2

⇒ 4d = 16

⇒ d = 4

From (i),

a = 7 – 2(4)

= -1

Also, we know sum of first 'n' terms of an AP is

Sum of first 20 terms is

= 10[-2 + 76]

= 740

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