Xavier Aptitude Test 2023 Paper

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Question : 65 of 102

Marks: +1, -0

Consider

a_{n+1}=\;{1}/{1+\;{1}/{a_n}}

for

n=1,2, . . . ., 2008,2009

where

a_1=1

. Find the value of

a_1 a_2+a_2 a_3+

a_3 a_4+. . .+a_{2008} a_{2009}

$\;\;{2009}/{1000}$
$\;\;{2009}/{2008}$
$\;\;{2008}/{2009}$
$\;\;{6000}/{2009}$
$\;\;{2008}/{6000}$

Solution:

Given that

a_1=1 & a_{n+1}=\;{1}/{1+\;{1}/{a_n}}

a_2=\;{1}/{1+\;{1}/{1}}=\;{1}/{2}, a_3=\;{1}/{1+\;{1}/{ (\;{1}/{2}) }}=\;{1}/{3}, . . .

This implies,

a_n=\;{1}/{n}

.
Required value

=a_1 a_2+a_2 a_3+. . . . . . .+a_{2008} a_{2009}

\;=\;{1}/{1} × \;{1}/{2}+\;{1}/{2} × \;{1}/{3}+. . . . .+\;{1}/{2008} × \;{1}/{2009}

\;= (\;{1}/{1}-\;{1}/{2}) + (\;{1}/{2}-\;{1}/{3}) +. . . .+ (\;{1}/{2008}-\;{1}/{2009})

\;=1-\;{1}/{2009}

\;=\;{2008}/{2009}

The answer is option C.

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