Chapter 12.2, Problem 52AYU

In Problems 39-56, find each sum.

${\displaystyle \underset{n=1}{\overset{80}{\sum }}\left(\frac{1}{3}n+\frac{1}{2}\right)}$

To determine

To find: The sum of the given sequence.

Answer to Problem 52AYU

$1120$

Explanation of Solution

Given:

${\displaystyle \sum }_{n=1}^{80}\left(\frac{1}{3}n+\frac{1}{2}\right)$

Formula Used:

Sum of the first $n$ terms of an arithmetic sequence.

$S_n=\frac{n}{2}\left[a_1+a_n\right]$

Calculation:

From the given sequence, $a_1=\frac{5}{6},n=\text{ 80 and }{a}_{80}=\frac{163}{6}$.

To find the sum ${\displaystyle \sum }_{n=1}^{80}\left(\frac{1}{3}n+\frac{1}{2}\right)$:

${\displaystyle \sum }_{n=1}^{80}\left(\frac{1}{3}n+\frac{1}{2}\right)=\frac{80}{2}\left[\frac{5}{6}+\frac{163}{6}\right]=40\times 28=1120$   (By Formula)