NIMCET 2016 Question Paper with solutions

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Question : 97 of 120

Marks: +1, -0

If

a_1,a_2

, ...

a_n

are in A.P. and

a_1

= 0 then the value

(a_3/a_2+a_4/a_3+...+a_n/a_{n-1})

-

a_2(1/a_2+1/a_1+...+1/a_{n-2})

is equal to :

(n - 2) + $1/(n-2)$
$1/{n-2}$
n - 2
n - $1/{n-2}$

Solution:

Given

a_1,a_2,...,a_n

are in AP and

a_1=0

∴a_2=d,a_3=2d

∵a_2=d,a_3=2d

∵(\;\;{a_{1}}/{a_{2}}+\;\;{a_{4}}/{a_{3}}+. . .+\;\;{a_{n}}/{a_{n-1}})-a_{2}(\;\;{1}/{a_{2}}+\;\;{1}/{a_{3}}+. . .+\;\;{1}/{a_{n-2}})

∴(\;\;{2d}/{d}+\;\;{dt}/{2dt}+\;\;{4}/{3}+...\;\;{(n-1)d}/{(n-1)d})-d(\;\;{1}/{d}+\;\;{1}/{2d}+...+\;\;{1}/{(n-3)d})

=[1+1+1+\;\;{1}/{2}+1+\;\;{1}/{3}+. . .+1+\;\;{1}/{n-2}]-[1+\;\;{1}/{2}+\;\;{1}/{3}+. . .+\;\;{1}/{n-3}]

=[1+\;\;{1}/{2}+\;\;{1}/{3}+. . .+\;\;{1}/{n-3}+\;\;{1}/{n-2}+n-2]-[1+\;\;{1}/{2}+\;\;{1}/{3}+. . .+\;\;{1}/{n-3}]

=(n-2)+1/{(n-2)}

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