Mathematics - Sequence & Series Question with Solution | TestHub

Let the common difference of an A.P. be $d$, then

$2a_7=a_5$   (given)

$\Rightarrow 2\left(a_1+6d\right)=a_1+4d$

$\Rightarrow a_1+8d=0 \dots \left(1\right)$

And,

$a_{11}=18$

$\Rightarrow a_1+10d=18 \dots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get

$a_1=-72,d=9$

So,

$a_{18}=a_1+17d=-72+153=81$

$a_{10}=a_1+9d=9$

Now,

$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\dots ..\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$

$=12\left(\frac{\sqrt{a_{11}}-\sqrt{a_{10}}}{ d}+\frac{\sqrt{a_{12}}-\sqrt{a_{11}}}{ d}+\dots \dots \frac{\sqrt{a_{18}}-\sqrt{a_{17}}}{ d}\right)$

$$\begin{gathered} =12\left(\frac{\sqrt{a_{18}}-\sqrt{a_{10}}}{d}\right) \\ =\frac{12\left(9-3\right)}{9}=8 \end{gathered}$$