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.

Given that,
\(S_7\) = 49
\(S_{17}\) = 289

We know, Sum of nth term;

\(\Rightarrow S_n = {{n} \over {2}} [2a + (n – 1)d]\)
\(\Rightarrow S_7= {{7} \over {2}} [2a + (n -1)d]\)
\(\Rightarrow S_7 = {{7} \over {2}} [2a + (7 -1)d]\)
\(\Rightarrow 49 ={{7} \over {2}} [2a + 6d]\)
\(\Rightarrow \) 7 = (a+3d)
\(\Rightarrow \) a + 3d = 7 …........ (i)

In the same way,

\(\Rightarrow S_{17} = {{17} \over {2}} [2a + (17-1)d]\)
\(\Rightarrow 289 = {{17} \over {2}} (2a + 16d)\)
\(\Rightarrow \) 17 = (a + 8d)
\(\Rightarrow \) a + 8d = 17 ……… (ii)

Subtracting equation (i) from equation (ii),
5d = 10
d = 2

From equation (i), we can write it as;
\(\Rightarrow \) a + 3(2) = 7
\(\Rightarrow \) a + 6 = 7
\(\Rightarrow \) a = 1

Hence,
\(\Rightarrow S_n = {{n} \over {2}} [2a + (n – 1)d]\)
= \({{n} \over {2}}[2(1)+(n – 1)×2]\) = \({{n} \over {2}}(2+2n-2)\) = \({{n} \over {2}}(2n)\) = \(n^2\)