Question

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.

Answer 1

Given that, \(S_7 = 49\) and \(S_{17} = 289\)
\(S_n = \frac n2 [2a + (n-1)d]\)

\(S_7 = \frac 72 [2a + (7-1)d]\)

\(49 = \frac {7}{2} (2a + 6d)\)
\(7 = (a + 3d)\)
\(a + 3d = 7 ……..(i)\)
Similarly, 
\(S_{17 }= \frac {17}{2} [2a + (17-1)d]\)

\(289 = \frac {17}{2} (2a + 16d)\)
\(17 = (a + 8d)\)
\(a + 8d = 17 …….(ii)\)
Subtracting equation \((i)\) from equation \((ii)\),
\(5d = 10\)
\(d = 2\)
From equation (i),
\(a + 3(2) = 7\)
\(a + 6 = 7\)
\(a = 1\)
\(Sn = \frac n2 [2a + (n-1)d]\)

\(Sn= \frac n2 [2(1) + (n-1)(2)]\)

\(S_n= \frac n2 (2 + 2n - 2)\)

\(S_n= \frac n2 (2n)\)

\(S_n= n^2\)