Chapter 11.3, Problem 53PPS

To determine

Find $a_1$ in a geometric series for which ${\text{S}}_{\text{n}}=1330;a_n=486;r=\frac{3}{2}$

Answer to Problem 53PPS

  $a_1=64$

Explanation of Solution

Given:

Find $a_1$ in a geometric series for which ${\text{S}}_{\text{n}}=1330;a_n=486;r=\frac{3}{2}$

Concept Used:

The general summation formula for GP series is ${\text{S}}_{\text{n}}=\frac{a_1-a_n·r}{1-r}$

Calculation:

  ${\text{S}}_{\text{n}}=1330;a_n=486;r=\frac{3}{2};\text{Find}{a}_1$

  $\begin{array}{l}{\text{S}}_{\text{n}}=\frac{a_1-a_n·r}{1-r}\\ \text{1330}=\frac{a_1-486·\frac{3}{2}}{1-\frac{3}{2}}=\frac{a_1-729}{-\frac{1}{2}}\\ -\frac{1}{2}·\text{1330}=-\frac{1}{2}·\frac{a_1-729}{-\frac{1}{2}}\\ -665=a_1-729\\ -665+729=a_1-729+729\\ a_1=64\end{array}$

Thus, $a_1=64$