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-n-

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

I F a_{1,} a_{2}, \dots . ., 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} + \dots + 1 / a_{n} . a_{1} =

2 / a_{1} + a_{n} [1 / a_{1} + 1 / a_{2} + \dots . + 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}} + \dots \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