If the third term of an A.P. is 5 and the ratio of its 6th term to the 10th term is 7 : 13, then find the sum of first 20 terms of this A.P.

Given, a₃ = 5
∴ 5 = a + 2d      (By formula a_n = a + (n - 1)d )
⇒ a = 5 - 2d. (Eq 1)

Also, $\dfrac{a$6}{a{10}} = \dfrac{7}{13}

$\Rightarrow \dfrac{a + 5d}{a + 9d} = \dfrac{7}{13} \\[1em] \Rightarrow 13(a + 5d) = 7(a + 9d) \\[1em] \Rightarrow 13a + 65d = 7a + 63d \\[1em] \Rightarrow 13a - 7a = 63d - 65d \\[1em] \Rightarrow 6a = -2d \\[1em]$

Putting value of a from Eq 1 in above equation

$\Rightarrow 6(5 - 2d) = -2d \\[1em] \Rightarrow 30 - 12d = -2d \\[1em] \Rightarrow 12d - 2d = 30 \\[1em] \Rightarrow 10d = 30 \\[1em] \therefore d = 3 \\[1em] a = 5 - 2d \\[1em] = 5 - 2(3) \\[1em] = 5 - 6 \\[1em] = -1.\\[1em] \therefore a = -1 \\[2em] S$n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S{20} = \dfrac{20}{2}[2 \times (-1) + (20 - 1) \times 3] \\[1em] = 10[-2 + 19 \times 3] \\[1em] = 10[-2 + 57] \\[1em] = 10 \times 55 \\[1em] = 550.

Hence, the sum of first 20 terms of the A.P. is 550.