Question

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

Answer 1

We know, n^th term of an AP is an = a + (n – 1) where ‘a’ and ‘d’ are first term and common difference of AP respectively 

Given, third term of an A.P. is 

7a₃ = 7

⇒ a + 2d = 7 (i)the seventh term exceeds three times the third term by 2.

⇒ a₇ = 3a₃ + 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 S_n = \(\frac{n}{2}[2a + (n-1)d]\) 

Sum of first 20 terms is

S₂₀ = \(\frac{20}{2}[2(-1)+19(4)]\)

= 10[-2 + 76] = 740