Geometric Progression And Sum Of GP

Geometric progression – Introduction:

In an A.P, the difference between \(n^{th}\) term and \((n-1)^{th}\) term will be a constant which is known as the common difference of the A.P.

An Arithmetic Progression is in the form,

\(a, a+d, a+2d, a+3d ….. a + (n-1)d\)

Now, what if the ratio of \(n^{th}\) term to \((n-1)^{th}\) term in a sequence is a constant?

For example, consider the following sequence,

\(2, 4, 8, 16, 32,……..\)

You can see that,

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

Similarly,

Cosider 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.

General term of a Geometric Progression:

We had learned to find the \(n^{th}\) term of an A.P, which was

\(a_n \)= \( a + (n-1)d\)

Similarly, in case of G.P,

Let a be the first term and r be the common ratio,

Then the second term, \(a_2 = a \times r = ar \)

Third term, \(a_{3} = a_{2} \times r = ar \times r = ar^{2}\)

Similarly, \(n^{th}\) term, \(a_{n} \) = \(ar^{n-1}\)

The terms of a finite G.P can be written as \(a, ar, ar^2, ar^3,……ar^{n-1}\)

Terms of an infinite G.P can be written as \(a, ar, ar^{2}, ar^{3}, ……ar^{n-1},…….\)

\(a + ar + ar^2 + ar^3 + ⋯ + ar^n\) is called finite geometric series.

\( a + ar + ar^2 + ar^3 + ⋯ + ar^n + ⋯ \) is called infinite geometric series.

Example: 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: \(5^{th}\) 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,

\(a×4^2\) = \(16\), \(a\)= \(1\)

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

=\( 1×4^7\) = \(16384\)

Sum of n terms of a 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

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

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

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

Example: Find the sum of 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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Practise This Question

If the sum of the series 1+2x+4x2+8x3+..... is a  finite number, then the complete solution of x will be - 
[UPSEAT 2002]