Question

If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ {a}_5=30 \\ \Rightarrow a+\left(5-1\right)d=30 \\ \Rightarrow a+4d=30...\left(\mathrm{i}\right) \\ \mathrm{Also},a_{12}=65 \\ \Rightarrow a+\left(12-1\right)d=65 \\ \Rightarrow a+11d=65.....\left(\mathrm{ii}\right) \\ \mathrm{Solving}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ 7d=35 \\ \Rightarrow d=5 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}d\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ a+4\times 5=30 \\ \Rightarrow a=10 \\ \therefore S_{20}=\frac{20}{2}\left[2\times 10+(20-1)\times 5\right] \\ \Rightarrow S_{20}=10\left[2\times 10+(20-1)\times 5\right] \\ \Rightarrow S_{20}=1150 \end{gathered}$$