Geometric Progression And Sum Of GP

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.

\(\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} = 2\)

Similarly,

Consider a series \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},……….\)

\(\frac{\frac{1}{2}}{1}\) = \(\frac{\frac{1}{4}}{\frac{1}{2}}\) = \(\large \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{\frac{1}{16}}{\frac{1}{8}} = \frac{1}{2}\)

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

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

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,

\(a,ar,ar^2,…..ar^{n-1}\)

Let \(S_n,a,r\) be the sum of n terms, first term and ratio of the G.P respectively.

Then, \(S_n\) = \(a + ar + ar^2 + ⋯ + ar^{n-1}\) —(1)

There are two cases, either \(r = 1\) or \(r ≠ 1\)

If r=1, then \(S_n\) = \( a + a + a + ⋯ a\) = \(na\)

When \(r ≠ 1\),

Multiply (1) with r gives,

\(rS_n\) = \( ar + ar^2 + ar^3 + ⋯ + ar^{n-1} + ar^n\)—(2)

Subtracting (1) from (2) gives

\(rS_n – S_n = (ar + ar^{2} + ar^{3} + …. + ar^{n-2} + ar^{n-1} + ar^{n}) – (a + ar + ar^{2} + …. + ar^{n-2} + ar^{n-1})\)

\((r – 1) S_n = ar^{n} – a = a(r^{n}-1)\)

\(S_n = a\frac{(r^{n}-1)}{(r – 1)} = a\frac{(1 – r^{n})}{(1 – r )}\)

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:

\(S_{\infty}=\sum_{n=1}^{\infty} ar^{n-1}=\frac{a}{1-r}; -1<r<1\)

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 × 0.1⁰) + (0.2 × 0.1¹) + (0.2 × 0.1²) + (0.2 × 0.1³) + ….∞

= (0.2) + (0.2 × 0.1) + (0.2 × 0.1²) + (0.2 × 0.1³) + ….∞

This of the form a + ar + ar² + ar³ + ….. ∞ (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.

Solved Examples

Example 1: If \(n^{th}\) 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

\(r\)= \(\frac{6}{3}\) = \(2\)

Since the first term, \(a\) = \(3\)

\(a_n\) = \(ar^{n-1}\)

\(192\) = \(3 \times 2^{n-1}\)

\(2^{n-1} = \frac{192}{3} = 64 = 2^6\)

\(n – 1 = 6 \)

n = 7

Therefore, 192 is \(7^{th}\) term of the G.P.

Example 2: \(5^{th}\) term and \(3^{rd}\) term of a G.P is 256 and 16 respectively. Find its \(8^{th}\) term.

Solution: \(ar^4\) = \(256\)—(1)

\(ar^2\) = \(16\)—(2)

Dividing (1) by (2) gives,

\(\frac{ar^4}{ar^2}\) = \(\frac{256}{16}\)

\(r^2\) = \(16\)

\(r\)= \(4\)

Substituting \(r\) = \(4\) in (2) gives,

\(a×4^2\) = \(16\), \(a\)= \(1\)

\(a_8\) = \(ar^7\)

=\( 1×4^7\) = \(16384\)

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

Solution: \(a\) = \(4\)

Common ratio,\(r = \frac{12}{4} = 3\)

\(n\) = \(6\)

Sum of n terms of a G.P,

\(S_n\) = \(\frac{a(r^n-1)}{(r-1)}\)

\(S_6\) = \(\frac{4(3^6-1)}{(3-1)}\)

=\(\frac{4(729-1)}{(2)}\) = \(2 × 728 \) = \(1456\)

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