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The sum to 50 terms of the series 1+$2\left( {\dfrac{{11}}{{50}}} \right) + 3{\left( {\dfrac{{11}}{{50}}} \right)^2} + .......$ is given by
A. 2500
B. 2550
C. 2450
D. none of these
Answer
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Hint-From the initial few terms given , first try to find out what would be the 50th term in the sequence and then find the sum to the 50 terms accordingly.
The given sequence is 1+ $2\left( {\dfrac{{11}}{{50}}} \right) + 3{\left( {\dfrac{{11}}{{50}}} \right)^2} + .......$
So, the sum to 50 terms of the series can be written as
${S_{50}} = 1 + 2\left( {\dfrac{{11}}{{50}}} \right) + 3{\left( {\dfrac{{11}}{{50}}} \right)^2} + .......50{\left( {\dfrac{{11}}{{50}}} \right)^{49}}$ ---------(i)
So, now as we see in the equation we have ${\left( {\dfrac{{11}}{{50}}} \right)^{n - 1}}$
Since , it is missing in the first term, we will multiply both the LHS and RHS by $\dfrac{{11}}{{50}}$
So, we get the equation as $\dfrac{{11}}{{10}}{S_{50}} = \dfrac{{11}}{{50}} + 2{\left( {\dfrac{{11}}{{50}}} \right)^2} + 3{\left( {\dfrac{{11}}{{50}}} \right)^3} + .......50{\left( {\dfrac{{11}}{{50}}} \right)^{50}}$-----(ii)
Now, let us subtract eq(i)-eq(ii)
So, we get ${S_{50}} - \dfrac{{11}}{{50}}{S_{50}} = 1 + \dfrac{{11}}{{50}} + {\left( {\dfrac{{11}}{{50}}} \right)^2} + {\left( {\dfrac{{11}}{{50}}} \right)^3} + .........{\left( {\dfrac{{11}}{{50}}} \right)^{49}} - 50{\left( {\dfrac{{11}}{{50}}} \right)^{50}}$
Now, as we can see the RHS is in geometric progression,
Because $\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{{T_3}}}{{{T_2}}} = \dfrac{{11}}{{50}}$
We know the sum of n terms of a geometric progression(G.P) is given by
${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}};$ if r<0
So, from this we get the equation as
$\dfrac{{39}}{{50}}{S_{50}}$ =$\dfrac{{1 - {{\left( {\dfrac{{11}}{{50}}} \right)}^{50}}}}{{1 - \left( {\dfrac{{11}}{{50}}} \right)}} - 50{\left( {\dfrac{{11}}{{50}}} \right)^{50}}$
Now, on taking LCM and shifting $\dfrac{{39}}{{50}}$ to the RHS and solving this further we can write this to be equal to
${S_{50}} = {\left( {\dfrac{{50}}{{39}}} \right)^2}\left[ {1 - {{\left( {\dfrac{{11}}{{50}}} \right)}^{50}}} \right] - \dfrac{{{{50}^2}}}{{39}}{\left( {\dfrac{{11}}{{50}}} \right)^{50}}$
Now on taking $\dfrac{{{{50}^2}}}{{39}}$ common from both the terms , we get the equation to be equal to
$\dfrac{{50}}{{39}}\left[ {\dfrac{{50}}{{39}} \times \dfrac{{39}}{{50}} \times {{\left( {\dfrac{{11}}{{50}}} \right)}^{49}} \times 11} \right] - 50 \times \dfrac{{11}}{{50}} \times {\left( {\dfrac{{11}}{{50}}} \right)^{49}}$
So, on solving this, we get
$50 \times 50 = 2500$
So, option A is the correct answer to this question
Note: In these types of questions, first try to bring the given series into a standard progression that is into AP, GP or HP and then solve. In this case, we have got a geometric progression, solve it in accordance to the data given
The given sequence is 1+ $2\left( {\dfrac{{11}}{{50}}} \right) + 3{\left( {\dfrac{{11}}{{50}}} \right)^2} + .......$
So, the sum to 50 terms of the series can be written as
${S_{50}} = 1 + 2\left( {\dfrac{{11}}{{50}}} \right) + 3{\left( {\dfrac{{11}}{{50}}} \right)^2} + .......50{\left( {\dfrac{{11}}{{50}}} \right)^{49}}$ ---------(i)
So, now as we see in the equation we have ${\left( {\dfrac{{11}}{{50}}} \right)^{n - 1}}$
Since , it is missing in the first term, we will multiply both the LHS and RHS by $\dfrac{{11}}{{50}}$
So, we get the equation as $\dfrac{{11}}{{10}}{S_{50}} = \dfrac{{11}}{{50}} + 2{\left( {\dfrac{{11}}{{50}}} \right)^2} + 3{\left( {\dfrac{{11}}{{50}}} \right)^3} + .......50{\left( {\dfrac{{11}}{{50}}} \right)^{50}}$-----(ii)
Now, let us subtract eq(i)-eq(ii)
So, we get ${S_{50}} - \dfrac{{11}}{{50}}{S_{50}} = 1 + \dfrac{{11}}{{50}} + {\left( {\dfrac{{11}}{{50}}} \right)^2} + {\left( {\dfrac{{11}}{{50}}} \right)^3} + .........{\left( {\dfrac{{11}}{{50}}} \right)^{49}} - 50{\left( {\dfrac{{11}}{{50}}} \right)^{50}}$
Now, as we can see the RHS is in geometric progression,
Because $\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{{T_3}}}{{{T_2}}} = \dfrac{{11}}{{50}}$
We know the sum of n terms of a geometric progression(G.P) is given by
${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}};$ if r<0
So, from this we get the equation as
$\dfrac{{39}}{{50}}{S_{50}}$ =$\dfrac{{1 - {{\left( {\dfrac{{11}}{{50}}} \right)}^{50}}}}{{1 - \left( {\dfrac{{11}}{{50}}} \right)}} - 50{\left( {\dfrac{{11}}{{50}}} \right)^{50}}$
Now, on taking LCM and shifting $\dfrac{{39}}{{50}}$ to the RHS and solving this further we can write this to be equal to
${S_{50}} = {\left( {\dfrac{{50}}{{39}}} \right)^2}\left[ {1 - {{\left( {\dfrac{{11}}{{50}}} \right)}^{50}}} \right] - \dfrac{{{{50}^2}}}{{39}}{\left( {\dfrac{{11}}{{50}}} \right)^{50}}$
Now on taking $\dfrac{{{{50}^2}}}{{39}}$ common from both the terms , we get the equation to be equal to
$\dfrac{{50}}{{39}}\left[ {\dfrac{{50}}{{39}} \times \dfrac{{39}}{{50}} \times {{\left( {\dfrac{{11}}{{50}}} \right)}^{49}} \times 11} \right] - 50 \times \dfrac{{11}}{{50}} \times {\left( {\dfrac{{11}}{{50}}} \right)^{49}}$
So, on solving this, we get
$50 \times 50 = 2500$
So, option A is the correct answer to this question
Note: In these types of questions, first try to bring the given series into a standard progression that is into AP, GP or HP and then solve. In this case, we have got a geometric progression, solve it in accordance to the data given
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