XAT 2023 Question Paper | Quantitative Aptitude and Data Interpretation

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Question 15 : Consider \(a_{n+1}=\frac{1}{1+\frac{1}{a_n}}\) for \(n=1,2, \ldots, 2008,2009\) where \(a_1=1\). Find the value of \(a_1 a_2+a_2 a_3+a_3 a_4+\cdots+a_{2008} a_{2009}\)

Video Explanation

Explanatory Answer

\( a _ { n + 1 } = \frac { 1 } { \frac { 1 } { a _ { n } } + 1 } \)
\( a _ { 1 } = 1 \)
\( a _ { 2 } = \frac { 1 } { 1 + 1 } = \frac { 1 } { 2 } \)
\( a _ { 3 } = \frac { 1 } { 2 + 1 } = \frac { 1 } { 3 } \)
Generalizing…, \( a _ { n } = \frac { 1 } { n + 1 } \)
\( = a _ { 1 } a _ { 2 } + a _ { 2 } a _ { 3 } + a _ { 3 } a _ { 4 } + \ldots + a _ { 2008 } a _ { 2009 } \) 
\( = \frac { 1 } { 1 } \cdot \frac { 1 } { 2 } + \frac { 1 } { 2 } \cdot \frac { 1 } { 3 } + \frac { 1 } { 3 } \cdot \frac { 1 } { 4 } + \ldots + \frac { 1 } { 2008 } \cdot \frac { 1 } { 2009 } \) 
\( = \frac { 1 } { 1 \times 2 } + \frac { 1 } { 2 \times 3 } + \frac { 1 } { 3 \times 4 } + \ldots + \frac { 1 } { 2008 \times 2009 } \)
\( = \frac { 2 - 1 } { 1 \times 2 } + \frac { 3 - 2 } { 2 \times 3 } + \frac { 4 - 3 } { 3 \times 4 } + \ldots + \frac { 2009 - 2008 } { 2008 \times 2009 } \) 
\( = \frac { 1 } { 1 } - \frac { 1 } { 2 } + \frac { 1 } { 2 } - \frac { 1 } { 3 } + \frac { 1 } { 3 } - \frac { 1 } { 4 } + \ldots + \frac { 1 } { 2008 } - \frac { 1 } { 2009 } \)
\( = \frac { 1 } { 1 } - \frac { 1 } { 2009 } \)
\( = \frac { 2008 } { 2009 } \)

The question is " Consider \(a_{n+1}=\frac{1}{1+\frac{1}{a_n}}\) for \(n=1,2, \ldots, 2008,2009\) where \(a_1=1\). Find the value of \(a_1 a_2+a_2 a_3+a_3 a_4+\cdots+a_{2008} a_{2009}\) "

Hence, the answer is '\(\frac{2008}{2009}\)'

Choice C is the correct answer.

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