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Expert-verified Found in: Page 187 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # What is the sum of the terms of the geometric progression a+ar+...+arn when ${\mathbf{r}}{\mathbf{\ne }}{\mathbf{1}}$.

The sum of the terms of the geometric progression is ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}+1}\right)}{1-\mathrm{r}}$

See the step by step solution

## Step 1: Geometric progression

Given:

${\mathrm{S}}_{\mathrm{n}}=\mathrm{a}+\mathrm{ar}+...+{\mathrm{ar}}^{\mathrm{n}}$ with $r\ne 1$

## Step 2: Finding sum of geometric progression

Let us determine ${\mathrm{rS}}_{\mathrm{n}}$:

${\mathrm{rS}}_{\mathrm{n}}=\mathrm{ar}+{\mathrm{ar}}^{2}+...+{\mathrm{ar}}^{\mathrm{n}+1}$

Next, we determine the difference between ${\mathrm{S}}_{\mathrm{n}}$ and ${\mathrm{rS}}_{\mathrm{n}}$

${S}_{n}-r{S}_{n}=\left(a+ar+\dots +a{r}^{n}\right)-\left(ar+a{r}^{2}+\dots +a{r}^{n+1}\right)\phantom{\rule{0ex}{0ex}}=a+ar+\dots +a{r}^{n}-ar-a{r}^{2}-\dots -a{r}^{n+1}\phantom{\rule{0ex}{0ex}}=a-a{r}^{n+1}$

Let us next use the distributive property on the obtained equality

${\mathrm{S}}_{\mathrm{n}}-{\mathrm{rS}}_{\mathrm{n}}=\mathrm{a}-{\mathrm{ar}}^{\mathrm{n}+1}\phantom{\rule{0ex}{0ex}}{\mathrm{S}}_{\mathrm{n}}\left(1-\mathrm{r}\right)=\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}+1}\right)$

Divide each side by 1-r

${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}+1}\right)}{1-\mathrm{r}}$

Thus, the sum of the terms of the geometric progression is $\frac{a\left(1-{r}^{n+!}\right)}{1-r}$ ### Want to see more solutions like these? 