Let T_r be the r^th term of a sequence for r = 1,2,3,..... such that  and T_r > 0  N. If sum of first n terms of this sequence is  and maximum value of T₅ is  where p & q are coprime numbers then value of (p + q) is-

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$$\begin{gathered} \mathrm{We}\mathrm{have} \\ \frac{{\mathrm{T}}_{\mathrm{r}+2}}{{\mathrm{T}}_{\mathrm{r}+1}}=\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{\mathrm{T}}_{\mathrm{r}}} \\ \mathrm{Suppose} \mathrm{we} \mathrm{take} \mathrm{r}=1 \\ \Rightarrow \frac{{\mathrm{T}}_3}{{\mathrm{T}}_2}=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1} \mathrm{i}.\mathrm{e}. \\ \frac{\mathrm{Third} \mathrm{Term}}{\mathrm{Second} \mathrm{Term}}=\frac{\mathrm{Second} \mathrm{Term}}{\mathrm{First} \mathrm{Term}} \mathrm{which} \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{Geometric} \mathrm{Progression}. \\ \mathrm{Hence}\mathrm{given}\mathrm{sequence}\mathrm{is}\mathrm{a}\mathrm{G}.\mathrm{P}. \\ \mathrm{Let}\mathrm{a}\mathrm{be}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{R}\mathrm{be}\mathrm{common}\mathrm{ratio} \\ {\mathrm{r}}_{\mathrm{th}} \mathrm{term} \mathrm{is} \mathrm{given} \mathrm{by} \\ {\mathrm{T}}_{\mathrm{r}}={\mathrm{aR}}^{\mathrm{r}-1} \\ \Rightarrow {\mathrm{T}}_6={\mathrm{aR}}^5 \\ \Rightarrow {\mathrm{T}}_5={\mathrm{aR}}^4 \\ \mathrm{Sum}\mathrm{of}\mathrm{n}\mathrm{terms}=\frac{\mathrm{a}\left({\mathrm{R}}^{\mathrm{n}}-1\right)}{\mathrm{R}-1} \\ \mathrm{According}\mathrm{to}\mathrm{question} \\ \frac{\mathrm{a}\left({\mathrm{R}}^{\mathrm{n}}-1\right)}{\mathrm{R}-1}=\left(1-{\left(\frac{{\mathrm{T}}_6}{{\mathrm{T}}_5}\right)}^{\mathrm{n}}\right) \\ \Rightarrow \frac{\mathrm{a}\left({\mathrm{R}}^{\mathrm{n}}-1\right)}{\mathrm{R}-1}=\left(1-{\left(\frac{{\mathrm{aR}}^5}{{\mathrm{aR}}^4}\right)}^{\mathrm{n}}\right) \\ \Rightarrow \frac{\mathrm{a}\left({\mathrm{R}}^{\mathrm{n}}-1\right)}{\mathrm{R}-1}=1-{\mathrm{R}}^{\mathrm{n}} \\ \Rightarrow \frac{\mathrm{a}\left({\mathrm{R}}^{\mathrm{n}}-1\right)}{\mathrm{R}-1}=-\left({\mathrm{R}}^{\mathrm{n}}-1\right) \\ \Rightarrow \mathrm{a}=1-\mathrm{R} \\ {\mathrm{T}}_5={\mathrm{aR}}^4 \\ =\left(1-\mathrm{R}\right){\mathrm{R}}^4 \\ {\mathrm{T}}_5={\mathrm{R}}^4-{\mathrm{R}}^5 \\ \mathrm{Differentiate}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{R} \\ \frac{{\mathrm{DT}}_5}{\mathrm{dR}}=4{\mathrm{R}}^3-5{\mathrm{R}}^4=5{\mathrm{R}}^3\left(\frac{4}{5}-\mathrm{R}\right) \\ \mathrm{Critical}\mathrm{points}\mathrm{are}\mathrm{R}=0\mathrm{and}\mathrm{R}=\frac{4}{5}\mathrm{and}\mathrm{method}\mathrm{of}\mathrm{interval}\mathrm{we}\mathrm{can}\mathrm{see} \\ \mathrm{that}\mathrm{R}=\frac{4}{5}\mathrm{is}\mathrm{point}\mathrm{maxima}. \\ {\mathrm{T}}_5=\left(1-\mathrm{R}\right){\mathrm{R}}^4 \\ {\left({\mathrm{T}}_5\right)}_{\max }=\left(1-\frac{4}{5}\right){\left(\frac{4}{5}\right)}^4 \\ =\frac{256}{3125}=\frac{\mathrm{p}}{\mathrm{q}} \\ \mathrm{p}+\mathrm{q}=256+3125=3381\left(\mathrm{Answer}\right) \\ \mathrm{If} \mathrm{you} \mathrm{have} \mathrm{any} \mathrm{specific} \mathrm{doubt} \mathrm{in} \mathrm{this} \mathrm{solution} \mathrm{or} \mathrm{answer} \mathrm{is} \mathrm{not} \mathrm{matching}, \\ \mathrm{feel} \mathrm{free} \mathrm{to} \mathrm{ask}. \end{gathered}$$

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