Chapter 11.7, Problem 5MRE

To determine

To calculate:

The sum of each series

  $5+9+13+\mathrm{.....}+45$

Answer to Problem 5MRE

The sum of each series is $275$ .

Explanation of Solution

Given information:

The given sequence is,

  $5+9+13+\mathrm{.....}+45$

Formula:

  $l=a+\left(n-1\right)d$

Where,

  $l$ : Last term

  $a:$ First term

  $d:$ Common difference

  $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Where,

  $a:$ First term

  $d:$ Common difference

  $S_n:$ Sum of numbers in the sequence

Calculation:

The given sequence is,

  $5+9+13+\mathrm{.....}+45$

Here the first term $a=5$ b and the common difference $d$ can be calculated as,

  $d=9-5=13-9=\mathrm{....}=4$

If there are $n$ terms in the above sequence which is in AP. Here the last term $l=45$ which is the $nth$ term and is given by the formula,

  $l=a+\left(n-1\right)d$

Now, put the values in the above formula,

  $\begin{array}{l}l=a+\left(n-1\right)d\\ 5+\left(n-1\right)\cdot 4-5=45-5\\ \left(n-1\right)\times 4=40\\ \frac{\left(n-1\right)\times 4}{4}=\frac{40}{4}\\ \left(n-1\right)=10\\ n-1+1=10+1\\ n=11\end{array}$

Now to calculate the sum of the series the formula is given as,

  $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Now, here $a=5,d=4,n=11$

Substitute all the values in the given formula.

  $\begin{array}{l}{S}_{11}=\frac{11}{2}\left[2\left(5\right)+\left(11-1\right)4\right]\\ =5.5\left[10+\left(10\right)\times 4\right]\\ =5.5\left[10+40\right]\\ =5.5\left[50\right]\\ =275\end{array}$