If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Let first term be a and common difference be d.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given,

Sum of first 7 terms = 49

$\therefore S_7 = \dfrac{7}{2}[2 \times a + (7 - 1)d] \\[1em] \Rightarrow 49 = \dfrac{7}{2}[2a + 6d] \\[1em] \Rightarrow 49 = \dfrac{7}{2} \times 2[a + 3d] \\[1em] \Rightarrow \dfrac{49}{7} = a + 3d \\[1em] \Rightarrow a + 3d = 7 \text{ …….(1)}.$

Given,

Sum of first 17 terms = 289

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

Subtracting equation (1) from (2), we get :

⇒ a + 8d - (a + 3d) = 17 - 7

⇒ a - a + 8d - 3d = 10

⇒ 5d = 10

⇒ d = $\dfrac{10}{5}$ = 2.

Substituting value of d in equation (1), we get :

⇒ a + 3 × 2 = 7

⇒ a + 6 = 7

⇒ a = 1.

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

Hence, sum of first n terms = n².