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Q14RE

Expert-verifiedFound in: Page 187

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+...+ar ^{n} when ${\mathbf{r}}{\mathbf{\ne}}{\mathbf{1}}$.**

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

Given:

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

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

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

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 $\frac{a\left(1-{r}^{n+!}\right)}{1-r}$

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