Question

The sum of first n terms of two APs are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms.

Answer 1

Let the first term of the first AP be a
And common difference be d
$$\begin{gathered} s_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ \mathrm{And} \\ {a}_n=a+\left(n-1\right)d \\ {a}_{12}=a+11d=3n+8 \\ \mathrm{Similarly},\mathrm{for}\mathrm{second}\mathrm{A}.\mathrm{P} \\ \mathrm{Let}\mathrm{first}\mathrm{term}\mathrm{be}A \\ \mathrm{And}\mathrm{common}\mathrm{difference}\mathrm{be}D \\ {S}_n\mathit{=}\frac{n}{2}\left[2A\mathit{+}\left(n\mathit{-}\mathit{1}\right)D\right] \end{gathered}$$
$$\begin{gathered} \mathrm{And} \\ {A}_n\mathit{=}A\mathit{+}\left(n\mathit{-}\mathit{1}\right)D \\ {A}_{12}\mathit{=}A\mathit{+}11D=7n+15 \\ \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{ratio}\mathrm{of}12\mathrm{th}\mathrm{term} \\ \frac{a_{12}\mathit{}\mathrm{of}\mathrm{first}\mathrm{A}.\mathrm{P}}{A_{12}\mathrm{of}\mathrm{second}\mathrm{A}.\mathrm{P}}\mathit{=}\frac{a+11d}{A+11D} \\ \frac{s_n\mathit{}\mathrm{of}\mathrm{first}\mathrm{A}.\mathrm{P}}{S_n\mathrm{of}\mathrm{second}\mathrm{A}.\mathrm{P}}\mathit{=}\frac{3n+8}{7n+15} \end{gathered}$$

$$\begin{gathered} \frac{{\displaystyle \frac{n}{2}\left[2a+\left(n-1\right)d\right]}}{\frac{n}{2}\left[2A+\left(n-1\right)D\right]}=\frac{3n+8}{7n+15} \\ \frac{2a+\left(n-1\right)d}{2A+\left(n-1\right)D}=\frac{{\displaystyle 3n+8}}{{\displaystyle 7n+15}} \\ \Rightarrow \frac{a+\left({\displaystyle \frac{n-1}{2}}\right)d}{A+\left(\frac{n-1}{2}\right)D}=\frac{a+11d}{A+11D}=\frac{{\displaystyle 3n+8}}{{\displaystyle 7n+15}}.....\left(1\right) \\ \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\frac{a+11d}{A+11D} \end{gathered}$$
$$\begin{gathered} \frac{n-1}{2}=11 \\ \Rightarrow n=23 \\ \mathrm{Put}n=23in\left(1\right) \\ \frac{a+\left({\displaystyle \frac{22}{2}}\right)d}{A+\left(\frac{22}{2}\right)D}=\frac{69+8}{161+15} \\ \frac{a+11d}{A+11D}=\frac{77}{176} \\ \frac{a+11d}{A+11D}=\frac{7}{16} \\ \mathrm{Ratio}\mathrm{of}12\mathrm{th}\mathrm{term}\mathrm{is}7:16 \end{gathered}$$