Chapter 12.2, Problem 44AYU

In Problems 39-56, find each sum.

$2+5+8+\cdots +41$

To determine

To find: The sum of the given sequence.

Answer to Problem 44AYU

$301$

Explanation of Solution

Given:

$\text{2 }+\text{ 5 }+\text{ 8 }+\dots +\text{ 41}$

Formula Used:

If $a_1,a_n$ and $d$ are known, then $n$^th term can be found by the formula,

$a_n=a_1+\left(n-1\right)d$   (Formula 1)

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

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

Calculation:

To find $n$:

From the given sequence, $a_1=\text{ 2},a_n=\text{ 41 and }d=\text{ 3}$.

$\text{41 }=\text{ 2 }+\left(n-1\right)\text{3 }=\text{ 2 }+\text{ 3}n-\text{ 3 }=\text{ 3}n-\text{ 1}$   ( By Formula 1)

$\Rightarrow n=\text{ 14}$

To find the sum $S_{14}$:

$S_{14}=\frac{14}{2}\left[\text{2 }+\text{ 41}\right]=\text{ 7 }\times \text{ 43 }=\text{ 301}$   (By Formula 2)