Geometric Progression and Infinite GP Sum | NIMCET

Quick answer: A geometric progression (GP) has a constant ratio r between terms. With first term a, the nth term is $a_n = ar^{n - 1}$ and the sum of n terms is $S_n = \frac{a(r^n - 1)}{r - 1}$ for $r \ne 1$ . If $\lvert r\rvert < 1$ , the infinite GP has sum $S_\infty = \frac{a}{1 - r}$ . The geometric mean of positive a and b is $\sqrt{ab}$ .

Compare with the arithmetic progression, where terms change by a constant difference.

Key formulas

Quantity	Formula
Common ratio	$r = \frac{a_2}{a_1}$
nth term	$a_n = ar^{n - 1}$
Sum of n terms ( $r > 1$ )	$S_n = \frac{a(r^n - 1)}{r - 1}$
Sum of n terms ( $r < 1$ )	$S_n = \frac{a(1 - r^n)}{1 - r}$
Sum when $r = 1$	$S_n = na$
Infinite sum ( $\lvert r\rvert < 1$ )	$S_\infty = \frac{a}{1 - r}$

Geometric mean

The GM of positive a and b is $G = \sqrt{ab}$ .
Three non-zero numbers a, b, c are in GP if and only if $b^2 = ac$ .

Inserting n geometric means between a and b: the common ratio is

$r = \left(\frac{b}{a}\right)^{\frac{1}{n + 1}}$

and the product of the n means is $\left(\sqrt{ab}\right)^n$ .

Properties of a GP

Property	Statement
Multiplying by k	Still a GP, same r
Reciprocals	Still a GP, ratio $\frac{1}{r}$
Powers	$a_n^k$ form a GP with ratio $r^k$
Equidistant terms	$a_1a_n = a_2a_{n - 1} = \ldots$
Choosing 3 terms	$\frac{a}{r}$ , $a$ , $ar$
Logs of a positive GP	Form an AP

Recurring decimals

A recurring decimal is an infinite GP.

Example: $0.\overline{3} = 0.3 + 0.03 + \ldots = \frac{0.3}{1 - 0.1} = \frac{1}{3}$ .

Example: $0.\overline{27} = \frac{0.27}{1 - 0.01} = \frac{27}{99} = \frac{3}{11}$ .

Worked examples

Example 1: nth term

Find the 8th term of 3, 6, 12, ….

$a = 3$ , $r = 2$ : $a_8 = 3 \cdot 2^7 = 384$ .

Example 2: sum of n terms

Find $1 + 3 + 9 + \ldots$ to 6 terms.

$S_6 = \frac{3^6 - 1}{3 - 1} = \frac{728}{2} = 364$

Example 3: infinite sum

Find $8 - 4 + 2 - 1 + \ldots$ .

$a = 8$ , $r = -\frac{1}{2}$ : $S_\infty = \frac{8}{1 + \frac{1}{2}} = \frac{16}{3}$ .

Example 4: three terms in GP

Three numbers in GP have product 216 and sum 19. Find them.

Step 1: Take $\frac{a}{r}$ , $a$ , $ar$ . Product $a^3 = 216$ , so $a = 6$ .

Step 2: $\frac{6}{r} + 6 + 6r = 19$ gives $6r^2 - 13r + 6 = 0$ , so $r = \frac{3}{2}$ or $\frac{2}{3}$ .

Answer: 4, 6, 9.

Example 5: sum of a special series

Find $0.7 + 0.77 + 0.777 + \ldots$ to n terms.

Step 1: Write it as $\frac{7}{9}[(1 - 0.1) + (1 - 0.01) + \ldots]$ .

Step 2: This is $\frac{7}{9}\left[n - \frac{0.1(1 - 0.1^n)}{0.9}\right] = \frac{7}{9}\left[n - \frac{1}{9}\left(1 - 10^{-n}\right)\right]$ .

Summary table

Idea	Key result
nth term	$ar^{n - 1}$
Finite sum	$\frac{a(r^n - 1)}{r - 1}$
Infinite sum	$\frac{a}{1 - r}$ , $\lvert r\rvert < 1$
a, b, c in GP	$b^2 = ac$
GM	$\sqrt{ab}$

Exam tips

Check $\lvert r\rvert < 1$ before using the infinite sum formula.
Use $\frac{a}{r}$ , $a$ , $ar$ when the product of three terms is given.
Write "7 + 77 + 777 + …" type series in terms of $10^k - 1$ to turn them into GPs.

Next, study harmonic progression and the AM-GM-HM inequality, or see the NIMCET Algebra chapter.

Common mistakes

Using $ar^n$ for the nth term.
Applying $S_\infty$ when $\lvert r\rvert \ge 1$ .
Forgetting the sign of a negative ratio in infinite sums.
Taking the GM of numbers with opposite signs.

Frequently Asked Questions

What is the nth term of a GP?

aₙ = arⁿ⁻¹.

What is the sum of an infinite GP?

If |r| < 1, the sum is a/(1 − r).

What is the geometric mean of two numbers?

For positive a and b, the GM is √(ab).

When are three numbers in GP?

When the square of the middle number equals the product of the other two: b² = ac.

How do you convert a recurring decimal to a fraction?

Treat it as an infinite GP. For example, 0.272727… = 0.27/(1 − 0.01) = 3/11.