Let [1 x_1,1 x_2,...,1 x_n,,] (xi [ ] 0 for i = 1, 2, ..., n) be in A.P.

Question

Let $${1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\,$$ (xi $$ \ne $$ 0 for i = 1, 2, ..., n) be in A.P. such that x1=4 and x21 = 20. If n is the least positive integer for which ${x_n} > 50,$ then $$\sum\limits_{i = 1}^n {\left( {{1 \over {{x_i}}}} \right)} $$ is equal to :

Accepted Answer

$$ \because $$ $\,\,\,$${1 \over {{x_1}}},{1 \over {{x_2}}},{1 \over {{x_3}}},.....,{1 \over {{x_n}}}$$ are in A.P. x1 = 4 and x21 = 20 Let 'd' be the common difference of this A.P. $$\therefore\,\,\,$its 21st term =${1 \over {{x_{21}}}} = {1 \over {{x_1}}} + \left[ {\left( {21 - 1} \right) \times d} \right]$$ \Rightarrow $$ $\,\,\,$$ d = $${1 \over {20}}$$ $$ \times $$ $$\left( {{1 \over {20}} - {1 \over 4}} \right)$$ $$ \Rightarrow $$ d = - $${1 \over {100}}$$ Also xn > 50(given). $$\therefore\,\,\,$$ $${1 \over {{x_n}}} = {1 \over {{x_1}}} + \left[ {\left( {n - 1} \right) \times d} \right]$$ $$ \Rightarrow $$ $\,\,\,$xn =${{{x_1}} \over {1 + \left( {n - 1} \right) \times d \times {x_1}}}$$\therefore\,\,\,$${{{x_1}} \over {1 + \left( {n - 1} \right) \times d \times {x_1}}} > 50$$ \Rightarrow $$ $\,\,\,$$ $${4 \over {1 + \left( {n - 1} \right) \times \left( { - {1 \over {100}}} \right) \times 4}} > 50$$ $$ \Rightarrow $$ $\,\,\,$1 + (n$-$1)$ \times $($-$${1 \over {100}}$)$ \times $4 <${4 \over {50}}$$ \Rightarrow $$ $\,\,\,$$ - $${1 \over {100}}(n - 1) < - \[{{23} \over {100}}$$ $$ \Rightarrow $$ $\,\,\,$n$-$> 23$ \Rightarrow $n > 24 Therefore$\,\,\,$n = 25.$ \Rightarrow $$ $\,\,\,$$ $\sum\limits_{i = 1}^{25} {{1 \over {{x_i}}}} $=${{25} \over 2}\left[ {\left( {2 \times {1 \over 4}} \right) + \left( {25 - 1} \right) \times \left( { - {1 \over {100}}} \right)} \right]$=${{13} \over 4}$

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