The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first n terms.

Let the first term of A.P. be a and common difference = d.

Given, sum of first 7 terms of an A.P. is 49,

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 49 = \dfrac{7}{2}[2 \times a + (7 - 1)d] \\[1em] \Rightarrow 49 = \dfrac{7}{2}[2a + 6d] \\[1em] \Rightarrow 49 = 7(a + 3d) \\[1em] \Rightarrow a + 3d = 7 ……..(i)$

Given, sum of first 17 terms of an A.P. is 289,

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 289 = \dfrac{17}{2}[2 \times a + (17 - 1)d] \\[1em] \Rightarrow 289 = \dfrac{17}{2}[2a + 16d] \\[1em] \Rightarrow 289 = 17(a + 8d) \\[1em] \Rightarrow a + 8d = 17 ……..(ii)$

Subtracting (i) from (ii) we get,

⇒ a + 8d - (a + 3d) = 17 - 7

⇒ a - a + 8d - 3d = 10

⇒ 5d = 10

⇒ d = 2.

Substituting value of d in (i) we get,

⇒ a + 3(2) = 7

⇒ a + 6 = 7

⇒ a = 1.

$\text{Sum of n terms } = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{n}{2}[2(1) + (n - 1)2] \\[1em] = \dfrac{n}{2}[2 + 2n - 2] \\[1em] = \dfrac{n}{2} \times 2n \\[1em] = n^2.$

Hence, sum of n terms = n².