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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.

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Solution

## Let the first term of the first AP be a And common difference be d ${s}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\mathrm{And}\phantom{\rule{0ex}{0ex}}{a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}{a}_{12}=a+11d=3n+8\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\mathrm{for}\mathrm{second}\mathrm{A}.\mathrm{P}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{first}\mathrm{term}\mathrm{be}A\phantom{\rule{0ex}{0ex}}\mathrm{And}\mathrm{common}\mathrm{difference}\mathrm{be}D\phantom{\rule{0ex}{0ex}}{S}_{n}\mathit{=}\frac{n}{2}\left[2A\mathit{+}\left(n\mathit{-}\mathit{1}\right)D\right]$ $\mathrm{And}\phantom{\rule{0ex}{0ex}}{A}_{n}\mathit{=}A\mathit{+}\left(n\mathit{-}\mathit{1}\right)D\phantom{\rule{0ex}{0ex}}{A}_{12}\mathit{=}A\mathit{+}11D=7n+15\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{ratio}\mathrm{of}12\mathrm{th}\mathrm{term}\phantom{\rule{0ex}{0ex}}\frac{{a}_{12}\mathit{}\mathrm{of}\mathrm{first}\mathrm{A}.\mathrm{P}}{{A}_{12}\mathrm{of}\mathrm{second}\mathrm{A}.\mathrm{P}}\mathit{=}\frac{a+11d}{A+11D}\phantom{\rule{0ex}{0ex}}\frac{{s}_{n}\mathit{}\mathrm{of}\mathrm{first}\mathrm{A}.\mathrm{P}}{{S}_{n}\mathrm{of}\mathrm{second}\mathrm{A}.\mathrm{P}}\mathit{=}\frac{3n+8}{7n+15}$ $\frac{\frac{n}{2}\left[2a+\left(n-1\right)d\right]}{\frac{n}{2}\left[2A+\left(n-1\right)D\right]}=\frac{3n+8}{7n+15}\phantom{\rule{0ex}{0ex}}\frac{2a+\left(n-1\right)d}{2A+\left(n-1\right)D}=\frac{3n+8}{7n+15}\phantom{\rule{0ex}{0ex}}⇒\frac{a+\left(\frac{n-1}{2}\right)d}{A+\left(\frac{n-1}{2}\right)D}=\frac{a+11d}{A+11D}=\frac{3n+8}{7n+15}.....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\frac{a+11d}{A+11D}$ $\frac{n-1}{2}=11\phantom{\rule{0ex}{0ex}}⇒n=23\phantom{\rule{0ex}{0ex}}\mathrm{Put}n=23in\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{a+\left(\frac{22}{2}\right)d}{A+\left(\frac{22}{2}\right)D}=\frac{69+8}{161+15}\phantom{\rule{0ex}{0ex}}\frac{a+11d}{A+11D}=\frac{77}{176}\phantom{\rule{0ex}{0ex}}\frac{a+11d}{A+11D}=\frac{7}{16}\phantom{\rule{0ex}{0ex}}\mathrm{Ratio}\mathrm{of}12\mathrm{th}\mathrm{term}\mathrm{is}7:16\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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