Sequence And Series Question 348
Question: If $ a_1,\ a_2,\ a_3…….a_{n} $ are in A.P., where $ a_{i}>0 $ for all $ i $ , then the value of $ \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+ $ $ ……..+\frac{1}{\sqrt{{a_{n-1}}}+\sqrt{a_{n}}}= $
[IIT 1982]
Options:
A) $ \frac{n-1}{\sqrt{a_1}+\sqrt{a_{n}}} $
B) $ \frac{n+1}{\sqrt{a_1}+\sqrt{a_{n}}} $
C) $ \frac{n-1}{\sqrt{a_1}-\sqrt{a_{n}}} $
D) $ \frac{n+1}{\sqrt{a_1}-\sqrt{a_{n}}} $
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Answer:
Correct Answer: A
Solution:
As given $ a_2-a_1=a_3-a_2=…….=a_{n}-{a_{n-1}}=d $ Where $ d $ is the common difference of the given A.P. Also $ a_{n}=a_1+(n-1)d $ . Then by rationalising each term, $ \frac{1}{\sqrt{a_2}+\sqrt{a_1}}+\frac{1}{\sqrt{a_3}+\sqrt{a_2}}+….+\frac{1}{\sqrt{a_{n}}+\sqrt{{a_{n-1}}}} $ $ =\frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+…..+\frac{\sqrt{a_{n}}-\sqrt{{a_{n-1}}}}{a_{n}-{a_{n-1}}} $ $ =\frac{1}{d}{ \sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+……+\sqrt{a_{n}}-\sqrt{{a_{n-1}}} } $ $ =\frac{1}{d}{ \sqrt{a_{n}}-\sqrt{a_1} }=\frac{1}{d}( \frac{a_{n}-a_1}{\sqrt{a_{n}}+\sqrt{a_1}} ) $ $ =\frac{1}{d}{ \frac{(n-1)d}{\sqrt{a_{n}}+\sqrt{a_1}} }=\frac{n-1}{\sqrt{a_{n}}+\sqrt{a_1}} $ .
BETA
