Question

\[\left( {\frac{1}{{2 \cdot 5}}} \right) + \left( {\frac{1}{{5 \cdot 8}}} \right) + \left( {\frac{1}{{8 \cdot 11}}} \right) + ...100\] terms.
(a) \[\frac{{25}}{{160}}\]
(b) \[\frac{1}{6}\]
(c) \[\frac{{25}}{{151}}\]
(d) \[\frac{{25}}{{152}}\]

Answer 1

Hint: Use all the basic fundamentals of the sequence and series. According to the given question, series 2, 5, 8 ….. and 5,8,11…. are in arithmetic progression(A.P) because differences between the two adjacent terms are constant.

Complete step by step Solution:
As per the given question, the series is given up to the 100 terms in the form of sum. So, Let us assume that the series is given, is equal to the S. So we can write that,
\[\begin{array}{*{20}{c}}
  { \Rightarrow S}& = &{\left( {\frac{1}{{2 \cdot 5}}} \right) + \left( {\frac{1}{{5 \cdot 8}}} \right) + \left( {\frac{1}{{8 \cdot 11}}} \right) + ...}
\end{array}\left( {\frac{1}{{(3n - 1)(3n + 2)}}} \right)\] up to 100 terms.
As we can see that the series 2, 5, 8 and series 5, 8, 11 are in arithmetic progression (A.P). So we will find the \[{n^{th}}\]terms of both the series. So we can write,
In the series \[2,{\text{ }}5,{\text{ }}8{\text{ }} \ldots \ldots \]
\[ \Rightarrow \begin{array}{*{20}{c}}
  a& = &2
\end{array}\]and \[\begin{array}{*{20}{c}}
  d& = &{{T_2} - {T_1}}
\end{array}\]. So, \[\begin{array}{*{20}{c}}
  d& = &{5 - 2}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  d& = &3
\end{array}\]
Now we know that the \[{n^{th}}\] term of the series is,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {T_n}}& = &{a + (n - 1)d}
\end{array}\]
Now,
\[ \Rightarrow \begin{array}{*{20}{c}}
  {{T_n}}& = &{2 + 3(n - 1)}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {T_n}}& = &{3n - 1}
\end{array}\]
And for the series \[5,{\text{ }}8,{\text{ }}11......\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow a}& = &5
\end{array}\] and \[\begin{array}{*{20}{c}}
  d& = &3
\end{array}\]
Now \[{n^{th}}\]term of this series is,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {T_n}}& = &{5 + (n - 1)3}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {T_n}}& = &{3n + 2}
\end{array}\]
Therefore, we can write given series as,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {S_n}}& = &{\left( {\frac{1}{{2 \cdot 5}}} \right) + \left( {\frac{1}{{5 \cdot 8}}} \right) + \left( {\frac{1}{{8 \cdot 11}}} \right) + ...}
\end{array}\left( {\frac{1}{{(3n - 1)(3n + 2)}}} \right)\]
Now, multiply and divide the above series by 3. So we will get,
 \[\begin{array}{*{20}{c}}
  { \Rightarrow {S_n}}& = &{\frac{3}{3}\left[ {\frac{1}{{2 \cdot 5}} + \frac{1}{{5 \cdot 8}} + \frac{1}{{8 \cdot 11}} + ...\frac{1}{{(3n - 1)(3n + 2)}}} \right]}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {S_n}}& = &{\frac{1}{3}\left[ {\frac{{5 - 2}}{{2 \cdot 5}} + \frac{{8 - 5}}{{5 \cdot 8}} + \frac{{11 - 8}}{{8 \cdot 11}} + ...\frac{{(3n + 2) - (3n - 1)}}{{(3n - 1)(3n + 2)}}} \right]}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {S_n}}& = &{\frac{1}{3}\left[ {\frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{11}} + ...\frac{1}{{(3n - 1)}} - \frac{1}{{(3n + 2)}}} \right]}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {S_n}}& = &{\frac{1}{3}\left[ {\frac{1}{2} - \frac{1}{{(3n + 2)}}} \right]}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {{S_n}}& = &{\frac{n}{{2(3n + 2)}}}
\end{array}\]
Now we have to find the sum of the given series up to 100 terms. So the value of n will be 100.Therefore, we will get.
\[\begin{array}{*{20}{c}}
  { \Rightarrow {S_n}}& = &{\frac{{100}}{{2(3 \times 100 + 2)}}}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {S_n}}& = &{\frac{{25}}{{151}}}
\end{array}\]
Now, the final answer is \[\frac{{25}}{{151}}\].

Hence, the correct option is C.

Note: Numerators of the given series are in arithmetic progression (A.P). First of all, find the \[{n^{th}}\]term of the series which are in arithmetic progression (A.P). So, Find the sum of the series up to \[{n^{th}}\]term and after that put the value of \[\begin{array}{*{20}{c}}
  n& = &{100}
\end{array}\].