An algebra problem by Akshat Sharda

Algebra Level 3

If A 1 , A 2 , A 3 , … A_{1},A_{2},A_{3},\ldots is an arithmetic progression, then find the value of ∑ i = 1 2 n ( − 1 ) i ( A i + A i + 1 A i − A i + 1 ) \displaystyle \sum^{2n}_{i=1}(-1)^{i}\left( \dfrac{A_{i}+A_{i+1}}{A_{i}-A_{i+1}}\right) .

− 2 n -2n n + 2 n+2 2 n 2n 2 n − 1 2n-1

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Akshat Sharda by Akshat Sharda , 21, India

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2 solutions

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Prakhar Bindal
Nov 8, 2016

Sorry but this is a shortcut

Let n = 1 and let the AP Be 1,2,3.......

Putting values in summation you will get -2 and hence -2n from options

Classic JEE Style :P

2 Helpful 0 Interesting 0 Brilliant 0 Confused

(just a note!) my solution comes as a comment coz i entered 2n instead of minus by mistake

even shorter solution

A i − A i + 1 A_{i}-A_{i+1} is − d -d common difference.

numerator is a telescopic sum A 2 n + 1 − A 1 A_{2n+1}-A_{1}

so − 2 n -2n

Rohith M.Athreya - 4 years, 7 months ago

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I did the same mistake man. So frustrating

Shreyash Rai - 4 years, 7 months ago

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