Question

Let $a_1,a_2,\dots ,a_n$ be in A.P. If $a_5=2a_7$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\dots +\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to

Answer 1

Answer: 8

$2a_5=a_5(\text{ given })$
$2\left(a_1+6d\right)=a_1+4d$
$a_1+8d=0$....(1)
$a_1+10d=18$...(2)
$\text{ By }(1)\text{ and (2) we get }{a}_1=-72,d=9$
$a_{18}=a_1+17d=-72+153=81$
$a_{10}=a_1+9d=9$
$12\left(\frac{\sqrt{a_{11}}-\sqrt{a_{10}}}{d}+\frac{\sqrt{a_{12}}-\sqrt{a_{11}}}{d}+\dots \dots \frac{\sqrt{a_{18}}-\sqrt{a_{17}}}{d}\right)$
$12\left(\frac{\sqrt{a_{18}}-\sqrt{a_{10}}}{d}\right)=\frac{12(9-3)}{9}=\frac{12\times 6}{6}=8$