"Let $a_{1}, a_{2}, \ldots, a_{n}$ be in A.P. If $a_{5}=2 a_{7}$ and $a_{11}=18$, then $$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$$ is equal to ____________."

Question

Accepted Answer

"$a_{11}=18$

$$ \begin{aligned} & a+10 d=18 \\\\ & a_{5}=2 a_{7} \\\\ & a+4 d=2(a+6 d) \\\\ & a=-8 d \end{aligned} $$

(i) and (ii) $\Rightarrow a=-72, d=9$.

On rationalising the denominator, given expression

$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{11}}}{-d}+\frac{\sqrt{a_{11}}-\sqrt{a_{12}}}{-d}+\ldots+\frac{\sqrt{a_{17}}-\sqrt{a_{18}}}{-d}\right]$

$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{18}}}{-d}\right]$

$=12\left[\frac{\sqrt{a_{11}-d}-\sqrt{a_{11}+7 d}}{-d}\right]$

$=12\left[\frac{\sqrt{18-9}-\sqrt{18+63}}{-9}\right]$ $=12 \times \frac{2}{3}=8$"

Let $a_{1}, a_{2}, \ldots, a_{n}$ be in A.P. If $a_{5}=2 a_{7}$ and $a_{11}=18$, then

$$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$$ is equal to ____________.

Solution

About this question

Drill 25 more like these. Every day. Free.

Let $a_{1}, a_{2}, \ldots, a_{n}$ be in A.P. If $a_{5}=2 a_{7}$ and $a_{11}=18$, then $$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$$ is equal to ____________.

Solution

About this question

More Sequences and Series questions

Drill 25 more like these. Every day. Free.

Let $a_{1}, a_{2}, \ldots, a_{n}$ be in A.P. If $a_{5}=2 a_{7}$ and $a_{11}=18$, then

$$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$$ is equal to ____________.