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Sum of n Terms

If ar>0;∀ r...

Question

If $a_{r} > 0; \forall r, n \in N$ and $a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}$ are in A.P, then
$\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =$

A

n - 1

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B

\frac{n (a_{1} + a_{2 n})}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}}

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C

\frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}}

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D

n + 1

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Solution

The correct option is A $\frac{n (a_{1} + a_{2 n})}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}}$
$a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}$ are in A.P.
Let $d$ be the common difference, then
$S = \frac{2 n}{2} (a_{1} + a_{2 n})$
$\Rightarrow S = n (a_{1} + a_{2 n})$
Consider , $\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}$
$= \frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} \frac{\sqrt{a_{1}} - \sqrt{a_{2}}}{\sqrt{a_{1}} - \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} \frac{\sqrt{a_{2}} - \sqrt{a_{3}}}{\sqrt{a_{2}} - \sqrt{a_{3}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} \frac{\sqrt{a_{n}} - \sqrt{a_{n + 1}}}{\sqrt{a_{n}} - \sqrt{a_{n + 1}}}$
$= \frac{S (\sqrt{a_{1}} - \sqrt{a_{2}})}{- n d} + \frac{S (\sqrt{a_{2}} - \sqrt{a_{3}})}{- n d} + . . . + \frac{S (\sqrt{a_{n}} - \sqrt{a_{n + 1}})}{- n d}$
$= \frac{S}{- n d} (\sqrt{a_{1}} - \sqrt{a_{2}} + \sqrt{a_{2}} - \sqrt{a_{3}} + . . . . . . . - \sqrt{a_{n}} + \sqrt{a_{n}} - \sqrt{a_{n + 1}})$
$= \frac{S}{- n d} (\sqrt{a_{1}} - \sqrt{a_{n + 1}})$
$= \frac{(a_{1} + a_{2 n})}{d} (\sqrt{a_{n + 1}} - \sqrt{a_{1}})$
$= \frac{(a_{1} + a_{2 n})}{d (\sqrt{a_{n + 1}} - \sqrt{a_{1}})} (a_{n + 1} - a_{1})$
$= \frac{(a_{1} + a_{2 n})}{d (\sqrt{a_{n + 1}} + \sqrt{a_{1}})} n d$
$= \frac{n (a_{1} + a_{2 n})}{(\sqrt{a_{n + 1}} + \sqrt{a_{1}})}$

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Similar questions

a_{r} > 0, r \in N

a_{1}, a_{2}, a_{3}, . . ., a_{2 n}

are in A.P. then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

a_{r} > 0, r ϵ N

a_{1}

a_{2}

a_{3}

a_{2 n}

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

Given that a_i $>$ 0 and i belongs to a set of natural numbers. If $a_{1}, a_{2}, a_{3} . . . . . a_{2 n}$ are in AP, then find the value of $\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n} - 1}{{\sqrt{a}}_{2} + \sqrt{a_{3}}} . . . . . . . . . . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}$

a_{1,} a_{2,} a_{3,} . . . . . . . . a_{2 n + 1}

are in A.P. then

\frac{a_{2 n + 1} - a_{1}}{a_{2 n + 1} + a_{1}} + \frac{a_{2 n + 1} - a_{2}}{a_{2 n + 1} + a_{2}} + . . . . . . + \frac{a_{2 n + 1} - a_{n}}{a_{2 n + 1} + a_{n}}

a_{1}, a_{2}, a_{3}, . . . . a_{2 n + 1}

are in A.P. then

\frac{a_{2 n + 1} - a_{1}}{a_{2 n + 1} + a_{1}} + \frac{a_{2 n} - a_{2}}{a_{2 n} + a_{2}} + . . . + \frac{a_{n + 2} - a_{n}}{a_{n + 2} + a_{n}}

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