**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,

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,

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

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

Similarly,

\(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},……….\)\(\frac{\frac{1}{2}}{1}\) = \(\frac{\frac{1}{4}}{\frac{1}{2}}\) =

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

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, \(a_2\) = \(a×r\) = \(ar\)

Third term, \(a_3\) =\(a_2×r\) = \(ar×r\) = \(ar^2\)

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?

First, we have to find the common ratio

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

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

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

\(192\) = \(3×2^{n-1}\)

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

\(n – 1\) = \(6\), \(n\) = \(7\)

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.

\(ar^4\) = \(256\)—(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,

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

\(a\) = \(4\)

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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