JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 20)
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 ____________.
$$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$$ is equal to ____________.
Answer
8
Explanation
$a_{11}=18$
$$ \begin{aligned} & a+10 d=18 \\\\ & a_{5}=2 a_{7} \\\\ & a+4 d=2(a+6 d) \\\\ & a=-8 d \end{aligned} $$
(i) and (ii) $\Rightarrow a=-72, d=9$.
On rationalising the denominator, given expression
$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{11}}}{-d}+\frac{\sqrt{a_{11}}-\sqrt{a_{12}}}{-d}+\ldots+\frac{\sqrt{a_{17}}-\sqrt{a_{18}}}{-d}\right]$
$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{18}}}{-d}\right]$
$=12\left[\frac{\sqrt{a_{11}-d}-\sqrt{a_{11}+7 d}}{-d}\right]$
$=12\left[\frac{\sqrt{18-9}-\sqrt{18+63}}{-9}\right]$ $=12 \times \frac{2}{3}=8$
$$ \begin{aligned} & a+10 d=18 \\\\ & a_{5}=2 a_{7} \\\\ & a+4 d=2(a+6 d) \\\\ & a=-8 d \end{aligned} $$
(i) and (ii) $\Rightarrow a=-72, d=9$.
On rationalising the denominator, given expression
$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{11}}}{-d}+\frac{\sqrt{a_{11}}-\sqrt{a_{12}}}{-d}+\ldots+\frac{\sqrt{a_{17}}-\sqrt{a_{18}}}{-d}\right]$
$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{18}}}{-d}\right]$
$=12\left[\frac{\sqrt{a_{11}-d}-\sqrt{a_{11}+7 d}}{-d}\right]$
$=12\left[\frac{\sqrt{18-9}-\sqrt{18+63}}{-9}\right]$ $=12 \times \frac{2}{3}=8$
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