Question

15. ​If the first term of a G.P. is 5 and the sum of the first three terms is 31/5 then find the common ratio.


16. In a GP, the ratio of the sum of first 3 terms to the sum of first 6 terms is 125:152. Find the common ratio of the GP.

Answer 1

Dear student
16
$$\begin{gathered} \mathrm{Let}\mathrm{a}\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{common}\mathrm{ratio}\mathrm{be}\mathrm{r}. \\ \mathrm{Let}{\mathrm{S}}_3\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{first}3\mathrm{terms}\mathrm{and}{\mathrm{S}}_6\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{first}6\mathrm{terms}\mathrm{of}\mathrm{G}.\mathrm{P} \\ \mathrm{A}.\mathrm{T}.\mathrm{Q} \\ \frac{{\mathrm{S}}_3}{{\mathrm{S}}_6}=\frac{125}{152} \\ \Rightarrow \frac{{\displaystyle \frac{\mathrm{a}\left(1-{\mathrm{r}}^3\right)}{1-\mathrm{r}}}}{\frac{\mathrm{a}\left(1-{\mathrm{r}}^6\right)}{1-\mathrm{r}}}=\frac{125}{152} \\ \Rightarrow \frac{1-{\mathrm{r}}^3}{1^2-{\left({\mathrm{r}}^3\right)}^2}=\frac{125}{152} \\ \Rightarrow \frac{\cancel{1-{\mathrm{r}}^3}}{\cancel{\left(1-{\mathrm{r}}^3\right)}\left(1+{\mathrm{r}}^3\right)}=\frac{125}{152}\left[\mathrm{using}{\mathrm{x}}^2-{\mathrm{y}}^2=\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}-\mathrm{y}\right)\right] \\ \Rightarrow 152=125\left(1+{\mathrm{r}}^3\right) \\ \Rightarrow 152=125+125{\mathrm{r}}^3 \\ \Rightarrow 27=125{\mathrm{r}}^3 \\ \Rightarrow {\mathrm{r}}^3=\frac{27}{125} \\ \Rightarrow \mathrm{r}=\frac{3}{5} \end{gathered}$$
$$\begin{gathered} \mathrm{Note}:{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{1-\mathrm{r}} \\ \mathrm{where}\mathrm{a}\mathrm{is}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{r}\mathrm{is}\mathrm{the}\mathrm{common}\mathrm{ratio},\mathrm{where}\mathrm{r}\ne 1 \end{gathered}$$
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