Before going to learn how to find the sum of a given Geometric Progression, first know what a GP is in detail.

If in a sequence of terms, each succeeding term is generated by multiplying each preceding term with a constant value, then the sequence is called a **geometric progression**. (GP), whereas the constant value is called the common ratio. For example, 2, 4, 8, 16, 32, 64, … is a GP, where the common ratio is 2.

[latex]\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} = 2[/latex]

Similarly,

Consider a series [latex]1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},……….[/latex]

[latex]\frac{\frac{1}{2}}{1}[/latex] = [latex]\frac{\frac{1}{4}}{\frac{1}{2}}[/latex] = [latex]\large \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{\frac{1}{16}}{\frac{1}{8}} = \frac{1}{2}[/latex]

In the given examples, the ratio is a constant. Such sequences are called Geometric Progressions. It is abbreviated as G.P.

A sequence [latex]a_1,a_2,a_3,…….a_n,….[/latex] is a G.P, then [latex]\frac{a_{k+1}}{a_k} = r~~~~~~~~~~~~ [k>1][/latex]

Where r is a constant which is known as common ratio and none of the terms in the sequence is zero.

Now, learn how to add GP if there are n number of terms present in it.

**Also check:** Arithmetic Progression

## Sum of Nth terms of G.P.

Consider the G.P,

[latex]a,ar,ar^2,…..ar^{n-1}[/latex]

Let [latex]S_n,a,r[/latex] be the sum of n terms, first term and ratio of the G.P respectively.

Then, [latex]S_n[/latex] = [latex]a + ar + ar^2 + ⋯ + ar^{n-1}[/latex] —(1)

There are two cases, either [latex]r = 1[/latex] or [latex]r ≠ 1[/latex]

If r=1, then [latex]S_n[/latex] = [latex] a + a + a + ⋯ a[/latex] = [latex]na[/latex]

When [latex]r ≠ 1[/latex],

Multiply (1) with r gives,

[latex]rS_n[/latex] = [latex] ar + ar^2 + ar^3 + ⋯ + ar^{n-1} + ar^n[/latex]—(2)

Subtracting (1) from (2) gives

[latex]rS_n – S_n = (ar + ar^{2} + ar^{3} + …. + ar^{n-2} + ar^{n-1} + ar^{n}) – (a + ar + ar^{2} + …. + ar^{n-2} + ar^{n-1})[/latex]

[latex](r – 1) S_n = ar^{n} – a = a(r^{n}-1)[/latex]

[latex]S_n = a\frac{(r^{n}-1)}{(r – 1)} = a\frac{(1 – r^{n})}{(1 – r )}[/latex]

Sum of n terms
Sn = [latex]\frac{a(r^n – 1)}{r-1}[/latex]; Where r [latex]\neq[/latex] 1 |

## Sum To Infinity of GP

If the number of terms in a GP is not finite, then the GP is called infinite GP. The formula to find the sum to infinity of the given GP is:

[latex]S_{\infty}=\sum_{n=1}^{\infty} ar^{n-1}=\frac{a}{1-r}; -1<r<1[/latex]

Here,

S_{∞} = Sum of infinite geometric progression

a = First term

r = Common ratio

n = Number of terms

This formula helps in converting a recurring decimal to the equivalent fraction. This can be observed from the following example:

0.22222222… = 0.2 + 0.02 + 0.002 + 0.0002 + …..∞

= (0.2 × 0.1^{0}) + (0.2 × 0.1^{1}) + (0.2 × 0.1^{2}) + (0.2 × 0.1^{3}) + ….∞

= (0.2) + (0.2 × 0.1) + (0.2 × 0.1^{2}) + (0.2 × 0.1^{3}) + ….∞

This of the form a + ar + ar^{2} + ar^{3} + ….. ∞ (infinite GP) such that a = 0.2 and r = 0.1.

Thus, 0.22222222… = 0.2/(1 – 0.1)

= 0.2/0.9

= 2/9

Hence, the equivalent fraction of 0.22222222… is 2/9.

## Video Lesson

### Sum of Infinite Terms of G.P.

### Solved Examples

**Example 1**: If [latex]n^{th}[/latex] term of the G.P 3, 6, 12, …. is 192, then what is the value of n?

**Solution**: First, we have to find the common ratio

[latex]r[/latex]= [latex]\frac{6}{3}[/latex] = [latex]2[/latex]

Since the first term, [latex]a[/latex] = [latex]3[/latex]

[latex]a_n[/latex] = [latex]ar^{n-1}[/latex]

[latex]192[/latex] = [latex]3 \times 2^{n-1}[/latex]

[latex]2^{n-1} = \frac{192}{3} = 64 = 2^6[/latex]

[latex]n – 1 = 6 [/latex]

n = 7

Therefore, 192 is [latex]7^{th}[/latex] term of the G.P.

**Example 2**: [latex]5^{th}[/latex] term and [latex]3^{rd}[/latex] term of a G.P is 256 and 16 respectively. Find its [latex]8^{th}[/latex] term.

**Solution**: [latex]ar^4[/latex] = [latex]256[/latex]—(1)

[latex]ar^2[/latex] = [latex]16[/latex]—(2)

Dividing (1) by (2) gives,

[latex]\frac{ar^4}{ar^2}[/latex] = [latex]\frac{256}{16}[/latex]

[latex]r^2[/latex] = [latex]16[/latex]

[latex]r[/latex]= [latex]4[/latex]

Substituting [latex]r[/latex] = [latex]4[/latex] in (2) gives,

[latex]a×4^2[/latex] = [latex]16[/latex], [latex]a[/latex]= [latex]1[/latex]

[latex]a_8[/latex] = [latex]ar^7[/latex]

=[latex] 1×4^7[/latex] = [latex]16384[/latex]

**Example 3**: Find the sum of the first 6 terms of the G.P 4, 12, 36,…..

**Solution**: [latex]a[/latex] = [latex]4[/latex]

Common ratio,[latex]r = \frac{12}{4} = 3[/latex]

[latex]n[/latex] = [latex]6[/latex]

Sum of n terms of a G.P,

[latex]S_n[/latex] = [latex]\frac{a(r^n-1)}{(r-1)}[/latex]

[latex]S_6[/latex] = [latex]\frac{4(3^6-1)}{(3-1)}[/latex]

=[latex]\frac{4(729-1)}{(2)}[/latex] = [latex]2 × 728 [/latex] = [latex]1456[/latex]

To know more about the concepts of sequence and series download BYJU’S – The Learning App.

how to solve cube root?

Hi,

Please find the link to solve cube root here: https://byjus.com/maths/cube-root-of-numbers/

Thanks!

Please visit: https://byjus.com/maths/how-to-find-cube-root/

Thanks a lot this is really helpful