Find the value of \[\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}\]

Hint: Let us assume \[\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}\] as S. Now we have to find the value of S. Now we have to split the sum in to two parts. The first part shows numbers and the other part shows the expression of variable n. We know that the sum of n natural numbers is equal to \[\dfrac{n(n+1)}{2}\]. Now by using this formula, we can find the value of S.

Complete step by step answer:
From the question, it is given that we have to find the value of \[\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}\].
Let us assume \[\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}\] is equal to S.
\[\begin{align}
  & \Rightarrow S=\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}....\text{(1)} \\
 & \Rightarrow \text{S=(5+5+5+}.....\text{(100 terms))-}\left( \dfrac{2}{n}+\dfrac{4}{n}+\dfrac{6}{n}+.....(100\text{ terms)} \right) \\
 & \Rightarrow S=5(100)-\left( \dfrac{2+4+6+....(100\text{ terms)}}{n} \right) \\
 & \Rightarrow S=500-2\left( \dfrac{1+2+3+.....(100\text{ terms)}}{n} \right) \\
\end{align}\]
We know that the sum of n natural numbers is equal to \[\dfrac{n(n+1)}{2}\].
\[\Rightarrow S=500-2\left( \dfrac{\sum\limits_{n=1}^{100}{n}}{n} \right)\]
\[\Rightarrow S=500-2\left( \dfrac{\dfrac{100(100+1)}{2}}{n} \right)\]
\[\begin{align}
  & \Rightarrow S=500-2\left( \dfrac{100(101)}{2(100)} \right) \\
 & \Rightarrow S=500-2(101) \\
 & \Rightarrow S=500-202 \\
 & \Rightarrow S=298 \\
\end{align}\]
So, it is clear that the value of S is equal to 298.
So, the value of \[\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}\] is equal to 298.

Note: This problem can be solved alternatively.
From the question, it is given that we have to find the value of \[\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}\].
Let us assume \[\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}\] is equal to S.
\[\begin{align}
  & \Rightarrow S=\left( 5-\dfrac{2}{n} \right)+\left( 5-\dfrac{4}{n} \right)+\left( 5-\dfrac{6}{n} \right)+........(100\text{ terms)}....\text{(1)} \\
 & \Rightarrow \text{S=(5+5+5+}.....\text{(100 terms))-}\left( \dfrac{2}{n}+\dfrac{4}{n}+\dfrac{6}{n}+.....(100\text{ terms)} \right) \\
 & \Rightarrow S=5(100)-\left( \dfrac{2+4+6+....(100\text{ terms)}}{n} \right) \\
\end{align}\]
We know that if the first term of the A.P is equal to a and common difference is equal to d, then the sum of numbers of A.P is equal to \[\dfrac{n}{2}[2a+(n-1)d]\].
\[\begin{align}
  & \Rightarrow S=500-\left( \dfrac{\left( \dfrac{100}{2} \right)[2(2)+(100-1)(2)]}{100} \right) \\
 & \Rightarrow S=500-\left( \dfrac{50[4+198]}{100} \right) \\
\end{align}\]
\[\begin{align}
  & \Rightarrow S=500-\left( \dfrac{50(202)}{100} \right) \\
 & \Rightarrow S=399 \\
\end{align}\]