Let T_r be the r th term of a sequence, for r = 1,2,3, .. If $3{\mathrm{T}}_{\mathrm{r}+1}={\mathrm{T}}_{\mathrm{r}}$ and ${\mathrm{T}}_7=\frac{1}{243},$then the value of $\sum _{\mathrm{r}=1}^{\mathrm{\infty }} \left({\mathrm{T}}_{\mathrm{r}}\cdot {\mathrm{T}}_{\mathrm{r}+1}\right)$ is

$\begin{array}{l}\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{\mathrm{T}}_{\mathrm{r}}}=\frac{1}{3}\\ \text{ If }{\mathrm{T}}_1=\mathrm{a},{\mathrm{T}}_2=\frac{\mathrm{a}}{3},{\mathrm{T}}_3=\frac{\mathrm{a}}{3^2},\dots \\ \therefore {\mathrm{T}}_7=\frac{\mathrm{a}}{3^6}=\frac{1}{243}\\ \Rightarrow \frac{\mathrm{a}}{3^5\times 3}=\frac{1}{3^5}\Rightarrow \mathrm{a}=3\\ \therefore {\mathrm{T}}_{\mathrm{r}}\times {\mathrm{T}}_{\mathrm{r}+1}=\frac{\mathrm{a}}{3^{\mathrm{r}-1}}\times \frac{\mathrm{a}}{3^{\mathrm{r}}}=\frac{{\mathrm{a}}^2}{3^{2\mathrm{r}-1}}=\frac{3^2}{3^{2\mathrm{r}-1}}=\frac{1}{3^{2\mathrm{r}-3}}\\ \therefore \sum _{\mathrm{r}=1}^{\mathrm{\infty }} {\mathrm{T}}_{\mathrm{r}}\cdot {\mathrm{T}}_{\mathrm{r}+1}=\left(3+\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+\dots \mathrm{\infty }\right)=\frac{3}{1-\frac{1}{3^2}}=\frac{27}{8}\\ \end{array}$