GP Formula (Nth term and sum) – Definition, Derivation, Examples

What is geometric progression?

Suppose you have a sequence of terms in front of your eyes, and you can get each succeeding term by multiplying the preceding term by a fixed number. We can also find the sum of a series if it is a geometric progression by GP formula. This sequence of numbers is a geometric progression, and the constant number is a called common ratio. For instance, 3, 9, 27, 81, ….. is a GP series where the common ratio is 3. 

So, if a sequence n_{1}, n_{2}, n_{3}, n_{4}, \ldots \ldots n_{x} is a GP series, then

\frac{n_{k}+1}{n_{k}}=\mathrm{r} \text {, } where k > 1.

Types of GP series

A geometric progression is of two types:

• Finite geometric progression – This type of GP series consists of a finite number of terms. For example, 1/2, 1/4, 1/8, 1/16 is a GP with 1/16 as the last term.
• Infinite geometric progression – This type of finite GP series consists of an infinite number of terms. For example, 5, 10, 20, 40, 80, ……. is an infinite GP series where the last term remains unknown.

The {\text n}^{\text {th }} term of a geometric sequence formula

To find the {\text n}^{\text {th }} term of a GP series, you must know the common ratio and the first term. If you do not know the common ratio, you can find it by calculating the ratio of two consecutive terms. The formula to find the nth term of a geometric progression is:

a_{n}=a r^{n-1}

where, 

a = the first term of the sequence

r = the common ratio of the sequence

and n = the number or the position of the unknown term.

Formula to find the sum of a GP series

The geometric sequence formula helps in calculating the sum of a geometric progression. Now, we know that there are two types of geometric progression. Hence there is a different formula to calculate the sum of each sequence.

Sum of finite geometric progression series 

If you have a fixed number of terms in a sequence then you can calculate the sum of finite GP formula easily.

\mathrm{S}=\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\mathrm{a}_{4}+\ldots \ldots+\mathrm{a}_{\mathrm{n}}

\mathrm{S}=\mathrm{a}_{1}+\mathrm{a}_{1} \mathrm{r}+\mathrm{a}_{1} \mathrm{r}^{2}+\mathrm{a}_{1} \mathrm{r}^{3}+\ldots \ldots+\mathrm{a}_{1} \mathrm{r}^{\mathrm{n}-1}                 – Equation 1

Multiplying both sides of equation 1 by r, we get

S r=a_{1} r+a_{1} r^{2}+a_{1} r^{3}+a_{1} r^{4}+\ldots \ldots+a_{1} r^{n}         – Equation 2

Subtracting equation 2 from equation 1,

S-S r=\left(a_{1}+a_{1} r+a_{1} r^{2}+a_{1} r^{3}+\ldots \ldots+a_{1} r^{n-1}\right)-\left(a_{1} r+a_{1} r^{2}+a_{1} r^{3}+a_{1} r^{4}+\ldots \ldots+a_{1} r^{n}\right)

S-S r=a_{1}-a_{1} r^{n}

Taking out common from both sides,

(1-r) S=a_{1}\left(1-r^{n}\right)

So,

S=\frac{a_{1}\left(1-r^{n}\right)}{1-r}

Therefore, the sum of finite GP formula when r < 1 is \frac{a_{1}\left(1-r^{n}\right)}{1-r}

But, if r > 1, then by subtracting equation 1 from equation 2, 

we get, \mathrm{S}=\frac{a_{1}\left(r^{n}-1\right)}{r-1}.

So, the sum of a finite GP series when r > 1 is, \mathrm{S}=\frac{a_{1}\left(r^{n}-1\right)}{r-1}.

Sum of infinite geometric progression series

When you do not know the total number of terms in a GP series, you must calculate the sum by considering the number of terms ‘n’ approaching to infinity ( ∞ ). The geometric progression formula sum to infinity is only applicable for the defined range of-

-1<\mathbf{r} \neq 0<+1

If r < 1, we can take

S=\frac{a_1 \cdot\left(1-r^{n}\right)}{1-r}

S=\frac{a_{1}-a_{1} r^{n}}{1-r}

\mathrm{S}=\frac{a_{1}}{1-r}-\frac{a_{1} r^{n}}{1-r}

Knowing that n → ∞, we get  \frac{a_{1} r^{n}}{1-r}\rightarrow 0

Thus, S=\frac{a_{1}}{1-r}

So, the sum of an infinite GP series is S=\frac{a_{1}}{1-r}

Solved Examples

Example 1:

Find the sum of 4 terms of the series, 2, 4, 8, 16.

Solution:

First, confirm that the given series is a GP. 

Divide each term by its preceding term and check if the common ratio is constant. In this case,

\frac{4}{2}= 2, \frac{8}{4}=2, \frac{16}{8}=2

Since the common ratio of these terms is 2, it is a geometric progression. 

So, r = 2

n = 4

since r > 1, we can use the formula, \mathrm{S}=\frac{a_{1}\left(r^{n}-1\right)}{r-1}

By substituting the above values in the equation, S=\frac{a_{1}\left(r^{n}-1\right)}{r-1}

S=\frac{2\left(2^{4}-1\right)}{2-1}

S = 2 x 15

S = 30

Example 2:

Find the 6^{\text {th }}term of a GP if the first term is 5 and the common ratio is 2.

Solution:

We have, r = 2, a = 5 and n = 6

By substituting these values in a_{n}=a r^{n-1}

a_{6}=5 \times 2^{6-1}

a_{6}=160

So, the 6th term of the GP is 160.