Question

What is the sum of the terms of the geometric progression a+ar+...+arⁿ when $\mathbf{r}\mathbf{\ne }\mathbf{1}$.

Short answer

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

Step-by-step solution

Geometric progression

Given:

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

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${\mathrm{rS}}_{\mathrm{n}}$

$$\begin{gathered} S_n-r{S}_n=\left(a+ar+\dots +a{r}^n\right)-\left(ar+a{r}^2+\dots +a{r}^{n+1}\right) \\ =a+ar+\dots +a{r}^n-ar-a{r}^2-\dots -a{r}^{n+1} \\ =a-a{r}^{n+1} \end{gathered}$$

Let us next use the distributive property on the obtained equality

$$\begin{gathered} {\mathrm{S}}_{\mathrm{n}}-{\mathrm{rS}}_{\mathrm{n}}=\mathrm{a}-{\mathrm{ar}}^{\mathrm{n}+1} \\ {\mathrm{S}}_{\mathrm{n}}(1-\mathrm{r})=\mathrm{a}(1-{\mathrm{r}}^{\mathrm{n}+1}) \end{gathered}$$

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