Chapter 9.2, Problem 56E

To determine

To find the partial sum of “ ${\displaystyle \sum _{n=51}^{100}7n}$ ”

Answer to Problem 56E

The partial sum of ${\displaystyle \sum _{n=51}^{100}7n}$ is $43925$

Explanation of Solution

Given information :

Given series ${\displaystyle \sum _{n=51}^{100}7n}$

As we have ${\displaystyle \sum _{n=51}^{100}7n}=7\left(51\right)+7\left(52\right)+7\left(53\right)+\mathrm{........}$ ,

For given arithmetic series, $a_1=7\left(51\right)$ and $d=7\left(52\right)-7\left(51\right)\Rightarrow d=7\left(52-51\right)=7$

So the $n^{th}$ term is $a_n=1+\left(n-1\right)\left(d\right)$ ,

  $\begin{array}{l}=7\left(51\right)+\left(n-1\right)\left(7\right)\\ =7\left(51+n-1\right)\\ =7\left(n+50\right)\end{array}$

  $\Rightarrow a_n=7\left(n+50\right)$

Therefore, $a_{51}=7\left(51+50\right)=7\left(101\right)=707$

  $\begin{array}{l}{a}_{100}=7\left(100+50\right)\\ =1050\end{array}$

To find the value of “ $n$ ” :

  $\begin{array}{l}\text{Since}{a}_{100}=1050\\ \Rightarrow a_{51}+(n-1)d=1050\\ \Rightarrow 707+(n-1)\left(7\right)=1050\\ \Rightarrow 707+7n-7=1050\end{array}$

  $\begin{array}{l}\Rightarrow 7n=1050-700\\ \Rightarrow n=\frac{350}{7}\\ \therefore n=50.\end{array}$

Thus, the partial sum of ${\displaystyle \sum _{n=51}^{100}7n}$ is $\left(\frac{50}{2}\right)\left(a_{51}+a_{100}\right)$

  $\begin{array}{l}{S}_{50}=\left(\frac{50}{2}\right)\left(707+1050\right)\\ =\left(25\right)\left(1757\right)\\ \therefore S_{50}=43925\end{array}$

Hence, the partial sum of ${\displaystyle \sum _{n=51}^{100}7n}$ is $43925$