Chapter 11.2, Problem 9CYU

To determine

Find the sum of the arithmetic series where $a_1=12;d=4,a_n=188;$

Answer to Problem 9CYU

Sum = 4500

Explanation of Solution

Given:

Given: $a_1=12;d=4,a_n=188;$

Concept Used:

The general formula for sum of arithmetic series is: ${\text{S}}_{\text{n}}=\frac{n}{2}[2a_1+(n-1)·d]$

If nth term is given, then the formula: ${\text{S}}_{\text{n}}=\frac{n}{2}(a_1+a_n)$

Calculation:

  $a_1=12;d=4,a_n=188;\text{find}{\text{S}}_{\text{n}}$

Now find the number of terms n:

  $\begin{array}{l}{a}_1=12;d=4,a_n=188;\\ a_n=a_1+(n-1)·d\\ 188=12+(n-1)·4\\ 188=12+4n-4\\ 188=4n+8\\ 188-8=4n+8-8\\ 180=4n\\ \frac{180}{4}=\frac{4n}{4}\\ n=45\end{array}$

  $\begin{array}{l}{a}_1=12;a_n=188,n=45;\text{find}{\text{S}}_{\text{n}}\\ {\text{S}}_{\text{n}}=\frac{n}{2}(a_1+a_n)\\ {\text{S}}_{\text{n}}=\frac{45}{2}(12+188)\\ {\text{S}}_{\text{n}}=45·100=4500\\ {\text{S}}_{\text{n}}=4500\end{array}$

Thus, the sum of the series is 4500.