Question

Let $T_r$ be the $r^{th}$ term of a sequence. If, for r = 1,2,3,.... . $3T_{r+1}=T_r$ and $T_7=\frac{1}{243},$ then the value of ${\sum }_{r=1}^{\infty }(T_r.T_{r+1})$ is

• A. $\frac{9}{2}$
• B. $\frac{27}{8}$
• C. $\frac{81}{8}$
• D. $\frac{81}{4}$

Answer 1

The correct option is B

$\frac{27}{8}$

$\frac{T_{r+1}}{T_r}=\frac{1}{3}$

If $T_1=a,T_2=\frac{a}{3},T_3=\frac{a}{3^2},.....$

$\therefore T_7=\frac{a}{3^6}=\frac{1}{243}\Rightarrow a=3$

$\therefore T_r.T_{r+1}=\frac{a}{3^{r-1}}.\frac{a}{3^r}=\frac{a^2}{3^{2r-1}}=\frac{3^2}{3^{2r-1}}$

$\Rightarrow {\sum }_{r=1}^{\infty }{T}_r.T_{r+1}=\frac{3^2}{3^{2r-1}}=3^{3-2r}$

$\Rightarrow {\sum }_{r=1}^{\infty }(T_r.T_{r+1})=(3+\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+.....\infty )=\frac{3}{1-\frac{1}{9}}=\frac{27}{8}$