JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 7)
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 :
$${1 \over 8}$$
3
$${{13} \over 8}$$
$${{13} \over 4}$$
Explanation
$$ \because $$$$\,\,\,$$ $${1 \over {{x_1}}},{1 \over {{x_2}}},{1 \over {{x_3}}},.....,{1 \over {{x_n}}}$$ 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.
$$ \Rightarrow $$$$\,\,\,$$$$\sum\limits_{i = 1}^{25} {{1 \over {{x_i}}}} $$ = $${{25} \over 2}\left[ {\left( {2 \times {1 \over 4}} \right) + \left( {25 - 1} \right) \times \left( { - {1 \over {100}}} \right)} \right]$$ = $${{13} \over 4}$$
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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