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Proof by mathematical induction

If a1,a2,a3...

Question

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

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

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B

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

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C

(n + 1) (a_{2} - a_{1})

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D

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

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Solution

The correct option is D $\frac{n (n + 1)}{2} \frac{a_{2} - a_{1}}{a_{n + 1}}$
$a_{1}, a_{2}, . . ., a_{2 n + 1}$ are in A.P
$\Rightarrow a_{2 n + 1} = a_{1} + 2 n d$
$a_{2 n} = a_{1} + (2 n - 1) d$
$. . . a_{2} = a_{1} + d$
$\Rightarrow a_{2 n + 1} - a_{1} = 2 a_{1} + 2 n d$
$a_{2 n} - a_{2} = 2 a_{1} + 2 n d$ and so on
Consider $\frac{a_{2 n + 1} - a_{1}}{a_{2 n + 1} + a_{1}} + \frac{a_{2 n} - a_{2}}{a_{2 n} + a_{2}} + \frac{a_{2 n - 1} - a_{3}}{a_{2 n - 1} + a_{3}} + . . . + \frac{a_{n + 2} - a_{n}}{a_{n + 2} + a_{n}}$
$= \frac{2 n d}{2 a_{1} + 2 n d} + \frac{(2 n - 2) d}{2 a_{1} + 2 n d} + . . . + \frac{2 d}{2 a_{1} + 2 n d}$
$= \frac{n d}{a_{1} + n d} + \frac{(n - 1) d}{a_{1} + n d} + . . . + \frac{d}{a_{1} + n d}$
$= \frac{d (1 + 2 + 3 + . . . + n)}{a_{1} + n d}$
$= \frac{d \frac{n (n + 1)}{2}}{a_{1} + n d}$
$= \frac{n (n + 1) d}{2 a_{n + 1}}$
$= \frac{n (n + 1) (a_{2} - a_{1})}{2 a_{n + 1}}$

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Similar questions

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}}

a_{r} > 0; \forall r, n \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 \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 $\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n} - 1}{{\sqrt{a}}_{2} + \sqrt{a_{3}}} . . . . . . . . . . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}$

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