IF a_(1,) a_2 ,.....,a_(n−1) ,a_n are in AP then prove that 1/a_1 .a_n + 1/a_2 .a_(n−1) + 1/a_3 .a_(n−2) +...+1/a_n .a_1 = 2/a_1 +a_(n ) [1/a_(1 ) +1/a_2 +....+1/a


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Question Number 27908 by v7277668420 last updated on 16/Jan/18

$I F a_{1,} a_{2}, . . . . ., a_{n - 1}, a_{n} a r e i n A P t h e n p r o v e t h a t$ $1 / a_{1} . a_{n} + 1 / a_{2} . a_{n - 1} + 1 / a_{3} . a_{n - 2} + . . . + 1 / a_{n} . a_{1} =$ $2 / a_{1} + a_{n} [1 / a_{1} + 1 / a_{2} + . . . . + 1 / a_{n}]$
Commented by v7277668420 last updated on 17/Jan/18

$p l e a s e h e l p$
	Answered by ajfour last updated on 17/Jan/18

	$l . h . s . = \underset{r = 1}{\sum^{n}} \frac{1}{a_{r} a_{n - r + 1}}$ $= \underset{r = 1}{\sum^{n}} \frac{1}{[a_{1} + (r - 1) d] [a_{1} + (n - r) d]}$ $= \frac{1}{2 a_{1} + (n - 1) d} \underset{r = 1}{\sum^{n}} \frac{[a_{1} + (r - 1) d] + [a_{1} + (n - r) d]}{[a_{1} + (r - 1) d] [a_{1} + (n - r) d]}$ $= \frac{1}{2 a_{1} + (n - 1) d} \underset{r = 1}{\sum^{n}} [\frac{1}{a_{1} + (n - r) d} + \frac{1}{a_{1} + (r - 1) d}]$ $= \frac{2}{a_{n} + a_{1}} [\frac{1}{a_{n}} + \frac{1}{a_{n - 1}} + \frac{1}{a_{n - 2}} + . . . \frac{1}{a_{1}}] .$
	Commented by v7277668420 last updated on 17/Jan/18

	$a n y e a s y m e t h o d$
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