# Sum of n terms of a GP

We will now discuss how to sum an arbitrary GP. Consider the sum of the first *n* terms of a GP with first term *a* and common ratio *r*:

\[S = a + ar + a{r^2} + ... + a{r^{n - 1}}\]

Multiply both sides by *r*, and write the terms with the same power of *r* below each other, as shown below:

\[\begin{align}S & = a + ar + a{r^2} + ... + a{r^{n - 1}}\\rS & = \,\,\,\,\,\,\,\,\,ar + a{r^2} + ... + a{r^{n - 1}} + a{r^n}\end{align}\]

Now, subtract the first relation from the second relation:

\[\begin{align}&\left( {r - 1} \right)S = a{r^n} - a\\ &\Rightarrow \,\,\,S = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\end{align}\]

This elegant artifice enables us to sum any GP, given *a* and *r*.

Let us take an example. Consider the following GP:

\[1,\,\,2,\,\,4,\,\,8,\,\,16,...\]

We have *a* = 1, and *r *=* *2. Now, let us find the sum of the first 12 terms of this GP. We have:

\[{S_{12}} = \frac{{\left( 1 \right)\left( {{2^{12}} - 1} \right)}}{{2 - 1}} = {2^{12}} - 1 = 4095\]

As another example, take the following GP:

\[\frac{1}{2},\,\,\frac{1}{6},\,\,\frac{1}{{18}},...\]

We have:

\[a = \frac{1}{2},\,\,\,\,r = \frac{1}{3}\]

Now, the sum of the first 6 terms of this GP will be:

\[\begin{align}&{S_6} = \frac{{\left( {\frac{1}{2}} \right)\left\{ {{{\left( {\frac{1}{3}} \right)}^6} - 1} \right\}}}{{\frac{1}{3} - 1}} &= \frac{{\frac{1}{2} \times \left( {\frac{1}{{729}} - 1} \right)}}{{ - \frac{2}{3}}}\\ &=\frac{{182}}{{243}}\end{align}\]

**Example 1:** In a GP, the sum of the first three terms is 16, and the sum of the next three terms is 128. Find the sum of the first *n* terms of the GP.

**Solution:** Let *a* and *r* be the first term and the common ratio of GP. We have:

\[\begin{array}{l}\left\{ \begin{array}{l}a + ar + a{r^2} = 16\\a{r^3} + a{r^4} + a{r^5} = 128\end{array} \right.\\ \Rightarrow \,\,\,\left\{ \begin{array}{l}a\left( {1 + r + {r^2}} \right) = 16\\a{r^3}\left( {1 + r + {r^2}} \right) = 128\end{array} \right.\end{array}\]

Dividing these two relations gives \({r^3} = 8\), or \(r = 2\). Substituting this in any of the two relations gives \(a = \frac{{16}}{7}\). The sum of the first *n* terms of the GP will be:

\[{S_n} = \frac{{\left( {\frac{{16}}{7}} \right)\left( {{2^n} - 1} \right)}}{{2 - 1}} = \frac{{16\left( {{2^n} - 1} \right)}}{7}\]

**Example 2:** For a GP, *a* is 5 and *r* is 2. The sum of a certain number of terms of this GP is 315. Find the number of terms and the last term.

**Solution:** If *n* is the number of terms, we have:

\[\begin{align}&{S_n} = \frac{{5\left( {{2^n} - 1} \right)}}{{2 - 1}} = 315\\ &\Rightarrow \,\,\,{2^n} - 1 = 63\,\,\, \Rightarrow \,\,\,n = 6\end{align}\]

The last term is

\[{T_6} = a{r^{n - 1}} = \left( 5 \right){\left( 2 \right)^5} = 160\]

**Example 3:** Consider the first *n* terms of a GP with first term *a *and common ratio *r*. *S* represents the sum of these terms, *P* their product, and *R* the sum of their reciprocals. Show that \({\left( {\frac{S}{R}} \right)^n} = {P^2}\).

**Solution: **We have:

\[\begin{align}&S = a + ar + a{r^2} + ... + a{r^{n - 1}}\\\,\,\,\, &\;\;= \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\\&R = \frac{1}{a} + \frac{1}{{ar}} + \frac{1}{{a{r^2}}} + ... + \frac{1}{{a{r^{n - 1}}}}\\\,\,\,\,\, &\;= \frac{{\left( {\frac{1}{a}} \right)\left( {\frac{1}{{{r^n}}} - 1} \right)}}{{\frac{1}{r} - 1}} = \frac{{{r^n} - 1}}{{a{r^{n - 1}}\left( {r - 1} \right)}}\end{align}\]

Thus,

\[\frac{S}{R} = {a^2}{r^{n - 1}}\]

Now, the product of the first *n* terms is:

\[\begin{align}&P = a \times ar \times a{r^2} \times ... \times a{r^{n - 1}}\\\,\,\,\,\, &\;\;= {a^n}{r^{\left( {1 + 2 + ... + \left( {n - 1} \right)} \right)}} = {a^n}{r^{\frac{{n\left( {n - 1} \right)}}{2}}}\end{align}\]

Finally, we see that:

\[{\left( {\frac{S}{R}} \right)^n} = {a^{2n}}{r^{n\left( {n - 1} \right)}} = {P^2}\]

**Example 4:** Find the sum of the following series:

\[5 + 55 + 555 + ...\,\,\,{\rm{to }}\,\, n\,\, {\rm{ terms}}\]

**Solution:** This series is not a GP, but by an interesting manipulation, we can convert it into a GP. Let *S* denote the required sum. We have:

\[\begin{align}&S = 5 + 55 + 555 + ...\,\,\,{\rm{to }} \,\, n\,\, {\rm{ terms}}\\\,\,\,\,&\;\;\;{\rm{ = 5}}\left( {1 + 11 + 111 + ...\,\,\,{\rm{to }}\,\, n\,\, {\rm{ terms}}} \right)\\\,\,\, &\;\;= \frac{5}{9}\left( {9 + 99 + 999 + ...\,\,\,{\rm{to }}\,\, n\,\, {\rm{ terms}}} \right)\\\,\,\, &\;= \frac{5}{9}\left\{ \begin{array}{l}\left( {10 - 1} \right) + \left( {{{10}^2} - 1} \right) + \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\, \,\,\left( {{{10}^3} - 1} \right) + ...\,\,\,{\rm{to }}\,\, n\,\, {\rm{ terms}}\end{array} \right\}\\\,\,\, &\;= \frac{5}{9}\left\{ {\left( {10 + {{10}^2} + {{10}^3} + ...\,\,\,{\rm{to }}\,\, n\,\, {\rm{ terms}}} \right) - n} \right\}\end{align}\]

Using the relation for the sum of the terms of a GP, we have:

\[\begin{align}&10 + {10^2} + {10^3} + ...\,\,\,{\rm{to }}\,\, n\,\, {\rm{ terms}}\\ &= \frac{{10\left( {{{10}^n} - 1} \right)}}{{10 - 1}} = \frac{{10\left( {{{10}^n} - 1} \right)}}{9}\end{align}\]

Thus,

\[\begin{align}&S = \frac{5}{9}\left\{ {\frac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right\}\\\,\,\,\, &\;= \frac{5}{{81}}\left( {{{10}^{n + 1}} - 9n - 10} \right)\end{align}\]