Question

if given series $\dfrac{1}{{{x_1}}},\dfrac{1}{{{x_2}}},.....,\dfrac{1}{{{x_n}}}$ (${x_i}$ $ \ne $ 0 for i = 1, 2, ……,n) be in A.P such that ${x_1}$ = 4 and ${x_{21}}$ = 20. If n is the least positive integer for which ${x_n}$ > 50, then $\sum\limits_{i = 1}^n {\left( {\dfrac{1}{{{x_i}}}} \right)} $ is equal to
A). 3
B). $\dfrac{{13}}{8}$
C). $\dfrac{{13}}{4}$
D). $\dfrac{1}{8}$

Answer 1

Hint:Before attempting this question one must have prior knowledge about A.P series and formula used in it and remember to use the formula of \[{x_n} = {x_1} + \left( {n - 1} \right)d\] for 21th term to find the common difference, use this information to approach the solution of the question.

Complete step-by-step solution:
According to the given information we have a A.P series i.e. \[\dfrac{1}{{{x_n}}},i = 1,2,3,...,n\] here ${x_1}$ = 4 and ${x_{21}}$ = 20
Now we have to find the common difference between the consecutive terms of this given A.P series
Using the formula of \[{x_n} = {x_1} + \left( {n - 1} \right)d\] to find the 21th term
\[\dfrac{1}{{{x_{21}}}} = \dfrac{1}{{{x_1}}} + \left( {21 - 1} \right)d\]
Now substituting the given values in the above equation we get
\[\dfrac{1}{{20}} = \dfrac{1}{4} + \left( {21 - 1} \right)d\]
$ \Rightarrow $\[\dfrac{1}{{20}} = \dfrac{1}{4} + 20 \times d\]
$ \Rightarrow $\[20 \times d = \dfrac{1}{{20}} - \dfrac{1}{4}\]
$ \Rightarrow $\[20 \times d = \dfrac{{1 - 5}}{{20}}\]
$ \Rightarrow $\[20 \times d = - \dfrac{1}{5}\]
$ \Rightarrow $\[d = - \dfrac{1}{{20 \times 5}}\]
$ \Rightarrow $\[d = - \dfrac{1}{{100}}\]
Using the formula for nth term i.e. \[{x_n} = {x_1} + \left( {n - 1} \right)d\]
For nth term in the given series
\[\dfrac{1}{{{x_n}}} = \dfrac{1}{{{x_1}}} + \left( {n - 1} \right)d\]
Simplifying the above equation we get
\[\dfrac{1}{{{x_n}}} = \dfrac{{1 + \left( {n - 1} \right) \times d \times {x_1}}}{{{x_1}}}\]
$ \Rightarrow $\[{x_n} = \dfrac{{{x_1}}}{{1 + \left( {n - 1} \right) \times d \times {x_1}}}\]
Since we know that ${x_n}$ > 50
Therefore \[\dfrac{{{x_1}}}{{1 + \left( {n - 1} \right) \times d \times {x_1}}} > 50\]
$ \Rightarrow $\[\dfrac{{{x_1}}}{{1 + \left( {n - 1} \right) \times d \times {x_1}}} > 50\]
Substituting the given values in the above equation we get
\[\dfrac{4}{{1 + \left( {n - 1} \right) \times \left( { - \dfrac{1}{{100}}} \right) \times 4}} > 50\]
$ \Rightarrow $\[\dfrac{4}{{50}} > 1 + \left( {n - 1} \right) \times \left( { - \dfrac{1}{{100}}} \right) \times 4\]
$ \Rightarrow $\[\dfrac{4}{{50}} - 1 > \left( {n - 1} \right) \times \left( { - \dfrac{1}{{100}}} \right) \times 4\]
$ \Rightarrow $\[\dfrac{{ - 46}}{{50}} > \left( {n - 1} \right) \times \left( { - \dfrac{1}{{100}}} \right) \times 4\]
$ \Rightarrow $\[\dfrac{{ - 23}}{{100}} > \left( {n - 1} \right) \times \left( { - \dfrac{1}{{100}}} \right)\]
$ \Rightarrow$ $- 23 > - (n – 1)$
$ \Rightarrow $ $24 < n $
Now we know that n should be greater than 24 therefore
$n = 25$
We have to find the value of $\sum\limits_{i = 1}^n {\left( {\dfrac{1}{{{x_i}}}} \right)} $ which is calculated by the formula $\sum\limits_{i = 1}^n {{x_i}} = \dfrac{n}{2}\left[ {2 \times {x_1} + \left( {n - 1} \right)d} \right]$
Substituting the values in the above formula we get
$\sum\limits_{i = 1}^{25} {\left( {\dfrac{1}{{{x_i}}}} \right)} = \dfrac{{25}}{2}\left[ {2 \times \dfrac{1}{4} + \left( {25 - 1} \right)\left( {\dfrac{{ - 1}}{{100}}} \right)} \right]$
$ \Rightarrow $$\sum\limits_{i = 1}^{25} {\left( {\dfrac{1}{{{x_i}}}} \right)} = \dfrac{{25}}{2}\left[ {\dfrac{1}{2} + \left( {24 \times \dfrac{{ - 1}}{{100}}} \right)} \right]$
$ \Rightarrow $$\sum\limits_{i = 1}^{25} {\left( {\dfrac{1}{{{x_i}}}} \right)} = \dfrac{{25}}{2} \times \dfrac{{26}}{{100}}$
$ \Rightarrow $$\sum\limits_{i = 1}^{25} {\left( {\dfrac{1}{{{x_i}}}} \right)} = \dfrac{{13}}{4}$
So the value of $\sum\limits_{i = 1}^n {\left( {\dfrac{1}{{{x_i}}}} \right)} $ is equal to $\dfrac{{13}}{4}$
Hence option C is the correct option.

Note: In the above question we used the concepts of arithmetic progression which can be explained as a sequence of numbers having a constant difference between the 2 consecutive numbers let’s take a help of an example to understand the concept of arithmetic progression suppose that you a sequence of numbers belong to whole numbers (0, 1, 2, 3,…… n) if you find the difference between the any 2 successive numbers like 3 and 2 is 1 same as for 2 and 1 the difference is 1 thus we can say that the sequence have same difference such sequences are called arithmetic progression.