Let \(S_n\) denote the sum of first \(n\) terms of an arithmetic progression. If \(S_{20}=790\) and \(S_{10}=145\), then…

Given, \(S_{20}=\frac{20}{2}[2 a+19 d]=790\)

\(2 a+19 d=79\)\(\ldots(1)\)

and, \(S_{10}=\frac{10}{2}[2 a+9 d]=145\)

\(2 a+9 d=29\)\(\ldots(2)\)

From (1) and (2)

\(a=-8, d=5\)

Now, \(S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d]\)

\(=\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20]\)

\(=405-10\)

\(=395\)