Let a 1 , a 2 , . . . . , a n be in A.P. If a 5 = 2 a 7 and a 11 = 18 , then 12 ( 1 a 10 + a 11 + 1 a 11 + a 12 + . . . + 1 a 17 + a 18 ) is equal to _______ . [2023]

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Solution

Q.
Let $a_{1}, a_{2}, . . . ., a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18,$ then $12 (\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + . . . + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}})$ is equal to _______ . [2023]

Ans.
(8)

We have, $a_{1}, a_{2}, \dots, a_{n}$ are in A.P.

$a_{5} = 2 a_{7}$

$\Rightarrow a + 4 d = 2 (a + 6 d) \Rightarrow a + 8 d = 0 ...(1)$

$and a_{11} = 18 \Rightarrow a + 10 d = 18 ...(2)$

$On solving (1) and (2), we get 2 d = 18 \Rightarrow d = 9 \Rightarrow a = - 72$

$a_{10} = a + 9 d = - 72 + 81 = 9$

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

$Now, 12 (\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \dots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{11}}}{{(\sqrt{a_{10}})}^{2} - {(\sqrt{a_{11}})}^{2}} + \frac{\sqrt{a_{11}} - \sqrt{a_{12}}}{{(\sqrt{a_{11}})}^{2} - {(\sqrt{a_{12}})}^{2}} + \dots + \frac{\sqrt{a_{17}} - \sqrt{a_{18}}}{{(\sqrt{a_{17}})}^{2} - {(\sqrt{a_{18}})}^{2}})$

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{11}}}{a + 9 d - a - 10 d} + \frac{\sqrt{a_{11}} - \sqrt{a_{12}}}{a + 10 d - a - 11 d} + \dots + \frac{\sqrt{a_{17}} - \sqrt{a_{18}}}{a + 16 d - a - 17 d})$

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

$= 12 (\frac{\sqrt{a_{10}} - \sqrt{a_{18}}}{- 9}) = \frac{12 \times (\sqrt{9} - \sqrt{81})}{- 9} = \frac{12 \times (3 - 9)}{- 9} = 8$

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