Question

The product of $t_5$ and $t_6$ of the progress $\frac{1}{4}$, $\frac{1}{2}$, 1, $\dots \dots$ is

• A. $t_8$
• B. $t_{11}$
• C. $t_{10}$
• D. $t_7$

Answer 1

The correct option is A $t_8$

The given sequence is a GP, with first terms $a=\frac{1}{4}$ and common ratio $r=\frac{t_2}{t_1}=\frac{\frac{1}{2}}{\frac{1}{4}}=2$
For a GP, the $nth$ term is given by $t_n=a{r}^{n-1}$
So, $t_5\times t_6=\left(a{r}^{5-1}\right)\times \left(a{r}^{6-1}\right)=(\frac{1}{4}\times 2^4)\times (\frac{1}{4}\times 2^5)$
$=\frac{1}{4}\times \left(2^4\right)\times (\frac{1}{4}\times 2^5)=\frac{1}{4}\times \left[\right(\frac{1}{4}\times 2^9]=\frac{1}{4}\times 2^7$
$=\frac{1}{4}\times 2^{8-1}=t_8$