Geometric Progression

ICSE Class 10 Mathematics — Chapter 11 Master Notes

1. Introduction to Sequences & Geometric Progression

Sequence: An arrangement of numbers written in a definite order according to a fixed mathematical rule.

Geometric Progression (G.P.): A sequence of non-zero numbers in which each term after the first is obtained by multiplying or dividing the preceding term by a fixed non-zero constant quantity.

Common Ratio ($r$): The fixed ratio between any term and its immediately preceding term.

$$r = \frac{t_2}{t_1} = \frac{t_3}{t_2} = \frac{t_4}{t_3} = \dots = \frac{t_n}{t_{n-1}}$$

2. General Term ($n^{\text{th}}$ Term) of a G.P.

Derivation of the $n^{\text{th}}$ Term Formula:

Let the first term of a G.P. be $a$ and its common ratio be $r$. Then:

$1^{\text{st}}$ term ($t_1$) $= a = a \cdot r^{1-1}$
$2^{\text{nd}}$ term ($t_2$) $= a \cdot r = a \cdot r^{2-1}$
$3^{\text{rd}}$ term ($t_3$) $= a \cdot r^2 = a \cdot r^{3-1}$
$4^{\text{th}}$ term ($t_4$) $= a \cdot r^3 = a \cdot r^{4-1}$

Continuing this pattern for $n$ terms:

General Term ($n^{\text{th}}$ Term from Beginning):

$$t_n = a \cdot r^{n-1}$$

Where $a = \text{first term}$, $r = \text{common ratio}$, and $n = \text{number of terms}$.

If a G.P. contains $n$ terms, the $n^{\text{th}}$ term is also the last term ($l$):

$$l = a \cdot r^{n-1}$$

$n^{\text{th}}$ Term from the End of a Finite G.P.:

If a finite G.P. has last term $l$ and common ratio $r$, the $n^{\text{th}}$ term from the end is:

$$t_n^{\text{end}} = \frac{l}{r^{n-1}} = l \cdot \left(\frac{1}{r}\right)^{n-1}$$

3. Selection of Terms in a G.P.

When the product of consecutive terms of a G.P. is given, select terms in the following symmetric forms to simplify algebra:

Number of Terms	Assumed Terms	Common Ratio
3 Terms	$\frac{a}{r}, \quad a, \quad ar$	$r$
4 Terms	$\frac{a}{r^3}, \quad \frac{a}{r}, \quad ar, \quad ar^3$	$r^2$
5 Terms	$\frac{a}{r^2}, \quad \frac{a}{r}, \quad a, \quad ar, \quad ar^2$	$r$

4. Important Properties of a G.P.

Key Algebraic Properties:

Equidistant Product Property: In a finite G.P., the product of terms equidistant from the beginning and end is constant and equal to the product of first and last terms. $$t_k \times t_{n-k+1} = a \times l$$
Three Consecutive Terms Condition: If $a, b, c$ are three consecutive terms of a G.P., then: $$\frac{b}{a} = \frac{c}{b} \implies b^2 = ac \implies b = \sqrt{ac}$$
Multiplication / Division Property: If each term of a G.P. is multiplied or divided by a non-zero constant, the resulting sequence is also a G.P. with the same common ratio.
Reciprocal Property: The reciprocals of the terms of a G.P. ($\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \dots$) also form a G.P. with common ratio $\frac{1}{r}$.
Logarithmic Relation with A.P.: If $a, b, c$ are positive numbers in G.P., then $\log a, \log b, \log c$ are in Arithmetic Progression (A.P.)!

5. Sum of $n$ Terms of a G.P. ($S_n$)

Derivation of Sum Formula ($S_n$):

Let $S_n = a + ar + ar^2 + ar^3 + \dots + ar^{n-1}$ --- (i)

Multiply (i) by $r$: $r \cdot S_n = ar + ar^2 + ar^3 + \dots + ar^n$ --- (ii)

Subtracting (ii) from (i):

$$S_n - r \cdot S_n = a - ar^n \implies S_n(1 - r) = a(1 - r^n)$$ $$\therefore S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \neq 1)$$

Sum Formulas for $S_n$:

When $r = 1$: $$S_n = n \cdot a$$
When $r < 1$: $$S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a - lr}{1 - r}$$
When $r > 1$: $$S_n = \frac{a(r^n - 1)}{r - 1} = \frac{lr - a}{r - 1}$$

Standard Board Summation Series ($5 + 55 + 555 + \dots$):

To evaluate $S_n = 5 + 55 + 555 + \dots$ up to $n$ terms:

$$S_n = 5(1 + 11 + 111 + \dots) = \frac{5}{9}(9 + 99 + 999 + \dots) = \frac{5}{9} \left[ (10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots \right]$$ $$S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{10 - 1} - n \right] = \frac{50}{81}(10^n - 1) - \frac{5n}{9}$$

6. Sum of Infinite Terms of a G.P. ($S_\infty$) & Recurring Decimals

Sum of Infinite Terms ($S_\infty$):

For an infinite G.P., if the numerical value of common ratio $|r| < 1$ (i.e. $-1 < r < 1$), then as $n \to \infty$, $r^n \to 0$.

$$S_\infty = \frac{a}{1 - r} \quad (|r| < 1)$$

Application to Recurring Decimals: Convert $0.\overline{437} = 0.4373737\dots$ into rational fraction form $\frac{p}{q}$.

Solution:

$$0.4373737\dots = 0.4 + 0.037 + 0.00037 + 0.0000037 + \dots$$ $$= \frac{4}{10} + \frac{37}{1000} \left[ 1 + \frac{1}{100} + \frac{1}{10000} + \dots \right]$$

The terms inside the bracket form an infinite G.P. with $a = 1, r = \frac{1}{100}$:

$$S_\infty = \frac{1}{1 - \frac{1}{100}} = \frac{100}{99}$$ $$\therefore \text{Fraction} = \frac{4}{10} + \frac{37}{1000} \times \frac{100}{99} = \frac{4}{10} + \frac{37}{990} = \frac{396 + 37}{990} = \frac{433}{990}$$

7. Geometric Mean (G.M.)

Single Geometric Mean between $a$ and $b$:

If $a, G, b$ are in G.P., then $G$ is called the Geometric Mean between $a$ and $b$.

$$\frac{G}{a} = \frac{b}{G} \implies G^2 = ab \implies G = \sqrt{ab}$$

Inserting $n$ Geometric Means ($G_1, G_2, \dots, G_n$) between $a$ and $b$:

Let $a, G_1, G_2, G_3, \dots, G_n, b$ form a G.P. containing total $(n + 2)$ terms.

First term $= a$, Last term $(n+2)^{\text{th}} = b$.
$b = a \cdot r^{(n+2)-1} = a \cdot r^{n+1} \implies r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$.
$G_1 = a \cdot r$, $G_2 = a \cdot r^2$, $\dots$, $G_n = a \cdot r^n$.

8. Step-by-Step Solved Master Examples & Interactive Chart

Interactive Comparison: G.P. Exponential Growth vs A.P. Linear Growth

Comparing G.P. ($a=2, r=2 \implies 2, 4, 8, 16, 32, 64$) vs A.P. ($a=2, d=2 \implies 2, 4, 6, 8, 10, 12$).

Example 1 (Finding Term Number): Which term of the G.P. $3, -6, 12, -24, \dots$ is $-384$?

Solution:

First term $a = 3$, common ratio $r = \frac{-6}{3} = -2$.

Let $t_n = -384 \implies a \cdot r^{n-1} = -384$:

$$3 \cdot (-2)^{n-1} = -384 \implies (-2)^{n-1} = -128$$ $$(-2)^{n-1} = (-2)^7 \implies n - 1 = 7 \implies n = 8$$

Answer: $-384$ is the $8^{\text{th}}$ term.

Example 2 (Finding G.P. from Two Given Terms): The $5^{\text{th}}$ term of a G.P. is $48$ and its $8^{\text{th}}$ term is $384$. Find the G.P.

Solution:

$t_5 = a r^4 = 48$ --- (i)

$t_8 = a r^7 = 384$ --- (ii)

Dividing (ii) by (i):

$$\frac{a r^7}{a r^4} = \frac{384}{48} \implies r^3 = 8 \implies r = 2$$

Substituting $r = 2$ in (i): $a(2^4) = 48 \implies 16a = 48 \implies a = 3$.

Answer: The G.P. is $3, 6, 12, 24, 48, \dots$

Example 3 (Term from the End): Find the $4^{\text{th}}$ term from the end of the G.P. $8, 4, 2, \dots, \frac{1}{128}$.

Solution:

Here common ratio $r = \frac{4}{8} = \frac{1}{2}$, last term $l = \frac{1}{128}$.

Using formula for $n^{\text{th}}$ term from end: $t_n^{\text{end}} = \frac{l}{r^{n-1}}$:

$$t_4^{\text{end}} = \frac{1/128}{(1/2)^{4-1}} = \frac{1/128}{(1/2)^3} = \frac{1/128}{1/8} = \frac{8}{128} = \frac{1}{16}$$

Example 4 (Infinite Sum Calculation): Find the sum to infinity of the G.P. $(\sqrt{2} + 1) + 1 + (\sqrt{2} - 1) + \dots$

Solution:

First term $a = \sqrt{2} + 1$. Common ratio $r = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1 < 1$.

Using $S_\infty = \frac{a}{1 - r}$:

$$S_\infty = \frac{\sqrt{2} + 1}{1 - (\sqrt{2} - 1)} = \frac{\sqrt{2} + 1}{2 - \sqrt{2}} = \frac{\sqrt{2} + 1}{\sqrt{2}(\sqrt{2} - 1)} = \frac{(\sqrt{2} + 1)^2}{\sqrt{2}(2 - 1)} = \frac{2 + 1 + 2\sqrt{2}}{\sqrt{2}} = \frac{4 + 3\sqrt{2}}{2}$$

9. Quick Formula & Revision Matrix

Concept	Given Input	Direct Formula
Common Ratio ($r$)	Terms $t_n, t_{n-1}$	$r = \frac{t_n}{t_{n-1}}$
$n^{\text{th}}$ Term ($t_n$)	$a, r, n$	$t_n = a \cdot r^{n-1}$
$n^{\text{th}}$ Term from End	$l, r, n$	$t_n^{\text{end}} = \frac{l}{r^{n-1}}$
Sum of $n$ Terms ($r > 1$)	$a, r, n$	$S_n = \frac{a(r^n - 1)}{r - 1} = \frac{lr - a}{r - 1}$
Sum of $n$ Terms ($r < 1$)	$a, r, n$	$S_n = \frac{a(1 - r^n)}{1 - r} = \frac{a - lr}{1 - r}$
Sum to Infinity ($S_\infty$)	$a, \|r\| < 1$	$S_\infty = \frac{a}{1 - r}$
Single Geometric Mean	Two numbers $a, b$	$G = \sqrt{ab}$

10. ICSE Board PYQs & High-Yield Practice Problems

ICSE Board PYQ 2015

BOARD The $4^{\text{th}}, 7^{\text{th}}$, and $10^{\text{th}}$ terms of a G.P. are $a, b, c$ respectively. Prove that $b^2 = ac$.

ICSE Board PYQ 2013

BOARD The $4^{\text{th}}$ term of a G.P. is $16$ and the $7^{\text{th}}$ term is $128$. Find the first term, common ratio, and the sum of its first 10 terms.

ICSE Board PYQ 2011

BOARD Find three numbers in G.P. whose sum is $21$ and the sum of their squares is $189$.

ICSE Board PYQ 2008

BOARD The sum of the first 3 terms of a G.P. is $\frac{13}{12}$ and their product is $-1$. Find the terms.

ICSE Board PYQ 2001

BOARD Insert three geometric means between $1$ and $256$.

ICSE Board PYQ 2000

BOARD The second term of a G.P. is $9$ and the sum of its infinite terms is $48$. Find its first term and common ratio.