Question

the sum of n terms of an AP whose first term is 6 and common difference is 40 is equal to 2n terms of another AP whose first term is 40 and common difference is 6. find n.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{a}\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{d}\mathrm{be}\mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{AP}. \\ \mathrm{Now},\mathrm{a}=6\mathrm{and}\mathrm{d}=40 \\ \mathrm{Let}\mathrm{A}\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{D}\mathrm{be}\mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{of}\mathrm{second}\mathrm{AP}. \\ \mathrm{A}=40;\mathrm{D}=6 \\ \mathrm{now},{\mathrm{S}}_{\mathrm{n}}={\mathrm{S}}_{2\mathrm{n}} \\ \Rightarrow \frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]=\frac{2\mathrm{n}}{2}\left[2\mathrm{A}+\left(2\mathrm{n}-1\right)\mathrm{D}\right] \\ \Rightarrow \left[2\times 6+\left(\mathrm{n}-1\right)40\right]=2\left[2\times 40+\left(2\mathrm{n}-1\right)6\right] \\ \Rightarrow 12+40\mathrm{n}-40=160+24\mathrm{n}-12 \\ \Rightarrow 40\mathrm{n}-24\mathrm{n}=160-12-12+40 \\ \Rightarrow 16\mathrm{n}=176 \\ \Rightarrow \mathrm{n}=11 \end{gathered}$$