Given,
\(S_7 = 49\) and \(S_{17} = 289\)
By using \(S_n = \frac n2 [2a + (n -1) d] \) we have,
\(S_7 = \frac 72 [ 2a +(7-1)d ] = 49\)
\(\Rightarrow 49 = \frac 72 [2a + 6d]\)
\(\Rightarrow \frac {49}{7} = a + 3d\)
\(\Rightarrow 7 = a + 3d\)
\(a + 3d = 7 \quad .....(i)\)
\(S_{17} = \frac {17}2 [2a + (17 - 1)d] = 289\)
\(\Rightarrow 289 = \frac {17}2 [2a + 16d] \)
\(\Rightarrow \frac{289}{17} = a + 8d\)
\(\Rightarrow 17= a + 8d\)
\(\Rightarrow a + 8d = 17 \quad.....(ii)\)
Substituting (i) from (ii), we get
\(5d = 10\)
\(d = \frac {10}5 = 2\)
\(\therefore d = 2\)
From eq (i),
\(a + 3d = 7\)
\(a + 3 \times 2= 7\)
\(a + 6 = 7\)
\(a = 7 - c\)
\(\therefore a = 1\)
\(\therefore S_{20} = \frac {20}2 [2 (1) + (20 - 1) (2)]\)
\(= 10 (2 + 19 \times 2)\)
\(= 10 (2 + 38)\)
\(= 10 \times 40\)
\(= 400\)
