If $ a_{1}, a_{2}, \ldots a_{n} $ are in arithmetic progression, where $ a_{1}0 $ for all $ i $. Then, $ \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} $ is equal to

Question

If $ a_{1}, a_{2}, \ldots a_{n} $ are in arithmetic progression, where $ a_{1}>0 $ for all $ i $. Then, $ \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} $ is equal to, $ \frac{n^{2}(n+1)}{2} $, $ \frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}} $, $ \frac{n(n-1)}{2} $, None of these

MARKS App · Accepted Answer

Since, a 1 , a 2 , … , a n are in arithmetic progression. Then, a 2 - a 1 = a 3 - a 2 = . . . = a n - a n - 1 = d Where d is common difference Now, 1 a 2 + a 1 + 1 a 3 + a 2 + … + 1 a n + a n - 1 = a 2 - a 1 d + a 3 - a 2 d + … + a n - a n - 1 d = 1 d ( a n - a 1 ) × a n + a 1 a n + a 1 = 1 d a n - a 1 a n + a 1 = n - 1 a n + a 1

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