Question

The sum of the first five terms of an AP is 55 and sum of first ten terms of an AP is 235, Find the sum of its first 20 terms.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{a}\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{d}\mathrm{be}\mathrm{the}\mathrm{common}\mathrm{difference}. \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{first}\mathrm{n}\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{AP}\mathrm{is}\mathrm{given}\mathrm{by} \\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ \mathrm{Now},{\mathrm{S}}_5=55 \\ \Rightarrow \frac{5}{2}\left[2\mathrm{a}+\left(5-1\right)\mathrm{d}\right]=55 \\ \Rightarrow 2\mathrm{a}+4\mathrm{d}=22 \\ \Rightarrow \mathrm{a}+2\mathrm{d}=11.....\left(1\right) \\ \mathrm{Now},{\mathrm{S}}_{10}=235 \\ \Rightarrow \frac{10}{2}\left[2\mathrm{a}+\left(10-1\right)\mathrm{d}\right]=235 \\ \Rightarrow 5\left(2\mathrm{a}+9\mathrm{d}\right)=235 \\ \Rightarrow 2\mathrm{a}+9\mathrm{d}=47.....\left(2\right) \\ \mathrm{Multiply}\left(1\right)\mathrm{by}2,\mathrm{we}\mathrm{get} \\ 2\mathrm{a}+4\mathrm{d}=22.....\left(3\right) \\ \mathrm{Subtracting}\left(3\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get} \\ 5\mathrm{d}=25 \\ \Rightarrow \mathrm{d}=5 \\ \mathrm{Now},\mathrm{put}\mathrm{d}=5\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{a}+2\times 5=11 \\ \Rightarrow \mathrm{a}=1 \\ \mathrm{Now},{\mathrm{S}}_{20}=\frac{20}{2}\left[2\times 1+\left(20-1\right)\left(5\right)\right] \\ =10\left(2+95\right) \\ =10\times 97 \\ =970 \end{gathered}$$