Mathematics - Sequence & Series Question with Solution | TestHub

Given 

$A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$  

Expanding using $A_n=A_1{r}^{n-1}$ formula we get,

 $\Rightarrow \left(A_1\right)\left(A_1{r}^2\right)\left(A_1{r}^4\right)\left(A_1{r}^6\right)=\frac{1}{1296}$

$\Rightarrow A_1^4{r}^{12}=\frac{1}{1296}\Rightarrow A_1{r}^3=\frac{1}{6} ...\left(\mathrm{i}\right)$

Also given ,

$A_2+A_4=\frac{7}{36}\Rightarrow A_{1}r+A_1{r}^3=\frac{7}{36}$

$\Rightarrow A_{1}r+\frac{1}{6}=\frac{7}{36}$

$\Rightarrow A_{1}r=\frac{1}{36} ...\left(\mathrm{ii}\right)$

Dividing equation (i) from equation (ii) we get, $r^2=6$ 

Now $A_6+A_8+A_{10}=A_1{r}^5+A_1{r}^7+A_1{r}^9$

$=A_1{r}^3\left[r^2+r^4+r^6\right]$ $=\frac{1}{6}\left[6+36+216\right]$  $=43$