Question

Sum of first 6 terms of aGP is 9 times the sum of the first 3terms .find common ratio

Answer 1

Dear student

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}:{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1},\mathrm{where}\mathrm{a}\mathrm{is}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{r}\mathrm{is}\mathrm{the}\mathrm{common}\mathrm{ratio} \\ \mathrm{and}\mathrm{r}\ne 1 \\ \mathrm{A}.\mathrm{T}.\mathrm{Q} \\ {\mathrm{S}}_6=9{\mathrm{S}}_3 \\ \Rightarrow \frac{{\mathrm{S}}_6}{{\mathrm{S}}_3}=\frac{9}{1} \\ \Rightarrow \frac{\frac{\mathrm{a}\left({\mathrm{r}}^6-1\right)}{\mathrm{r}-1}}{\frac{\mathrm{a}\left({\mathrm{r}}^3-1\right)}{\mathrm{r}-1}}=9 \\ \Rightarrow \frac{{\mathrm{r}}^6-1}{{\mathrm{r}}^3-1}=9 \\ \Rightarrow \frac{\left({\mathrm{r}}^3-1\right)\left({\mathrm{r}}^3+1\right)}{{\mathrm{r}}^3-1}=9 \\ \Rightarrow {\mathrm{r}}^3+1=9 \\ \Rightarrow {\mathrm{r}}^3=8 \\ \Rightarrow \mathrm{r}=2 \end{gathered}$$
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