QJEEM18Mab501: JEE Main 2018 Mathematics - Class 11 JEEM18Mab501 Solution: Let (x i 0 for i = 1, 2, ..., n) be in A.P. such that x 1 =4…

Question

Verified Solution • 2025–26 Session moderate

Let

{1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\,

(x_i

\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} &gt; 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 21^st 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 x_n > 50(given).

\therefore\,\,\,

{1 \over {{x_n}}} = {1 \over {{x_1}}} + \left[ {\left( {n - 1} \right) \times d} \right]

\Rightarrow

\,\,\,

x_n =

{{{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}}} &gt; 50

\Rightarrow

\,\,\,

{4 \over {1 + \left( {n - 1} \right) \times \left( { - {1 \over {100}}} \right) \times 4}} &gt; 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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