Question

Find the sum of 30 terms of an A.P. whose second term is 2 and seventh term is 22.
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Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{a}\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{d}\mathrm{be}\mathrm{the}\mathrm{common}\mathrm{difference}. \\ \mathrm{Now},{\mathrm{a}}_2=2 \\ \Rightarrow \mathrm{a}+\mathrm{d}=2....\left(1\right) \\ \mathrm{Now},{\mathrm{a}}_7=22 \\ \Rightarrow \mathrm{a}+6\mathrm{d}=22.....\left(2\right) \\ \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get} \\ 6\mathrm{d}-\mathrm{d}=22-2 \\ \Rightarrow 5\mathrm{d}=20 \\ \Rightarrow \mathrm{d}=\frac{20}{5}=4 \\ \mathrm{Put}\mathrm{d}=4\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{a}+4=2 \\ \Rightarrow \mathrm{a}=2-4=-2 \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{first}\mathrm{n}\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{AP}\mathrm{is}\mathrm{given}\mathrm{by} \\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ \mathrm{Now},{\mathrm{S}}_{30}=\frac{30}{2}\left[2\times \left(-2\right)+\left(30-1\right)\left(4\right)\right] \\ =15\left[-4+116\right] \\ =15\times 112 \\ =1680 \end{gathered}$$