ICSE Class 10 Maths Chapter 11 Geometric Progression

Geometric Progression

11.1 Introduction

A sequence is an arrangement of numbers written in a definite order according to a certain given rule.

(i) 2, 6, 18, 54, ... is a sequence in which each term multiplied by 3 gives its next term.

(ii) \(1, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, \ldots\) is a sequence in which each term multiplied by \(-\frac{1}{2}\) gives its next term.

(iii) 6, 4, 2, 0, ... is a sequence in which each term decreased by 2 gives its next term.

When the members (terms) of a sequence are connected using a positive/negative sign, we get a series.

Each of following is a series:

(i) \(2 + 6 + 18 + 54 + \ldots\)

(ii) \(1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \ldots\)

(iii) \(6 + 4 + 2 + 0 + \ldots\)

In general, a sequence and a series are considered to be the same.

11.2 Geometric Progression

A sequence, in which each of its terms can be obtained by multiplying or dividing its preceding term by a fixed quantity, is called a geometric progression and the fixed quantity is called common ratio.

In general, geometric (geometrical) progression is abbreviated as G.P. and common ratio is denoted by letter r.

(i) Sequence 2, 4, 8, 16, ... is a G.P. as each term of it multiplied by 2 gives its next term. [Clearly, common ratio \(r = 2\)].

(ii) Sequence \(6, 2, \frac{2}{3}, \ldots\) is a G.P. as each of its terms can be obtained by multiplying its preceding term by \(\frac{1}{3}\). [Clearly, common ratio \(r = \frac{1}{3}\)].

(iii) \(60, -24, 9\frac{3}{5}, -3\frac{21}{25}, \ldots\) is a G.P. with common ratio \(-\frac{2}{5}\).

In a G.P., common ratio is obtained by dividing any term of it by its preceding term.

For G.P. 16, 8, 4, 2, 1, \(\frac{1}{2}\), ...

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

\(\therefore \frac{8}{16} = \frac{4}{8} = \frac{2}{4} = \ldots = \frac{1}{2} =\) common ratio (r)

If first term of a G.P. is 5 and its common ratio is 3, then

G.P. = 5, 5 × 3, 5 × 3 × 3, 5 × 3 × 3 × 3, ...

= 5, 15, 45, 135, ...

Which can also be written as: 5 + 15 + 45 + 135 + ...

11.3 General Term Of A Geometric Progression

Let the first term of a geometric progression = a and its common ratio = r, then its:

first term, \(t_1 = a = ar^{1-1}\),

second term, \(t_2 = ar = ar^{2-1}\),

third term, \(t_3 = ar^2 = ar^{3-1}\),

fourth term, \(t_4 = ar^3 = ar^{4-1}\)

First term = a,
common ratio = r and
nth term = \(ar^{n-1}\)

and nth term, \(t_n = ar^{n-1}\)

If a geometric progression has n terms only, its nth term is called its last term which is denoted by letter l.

Last term (l) = \(ar^{n-1}\)

i.e.

nth term = \(ar^{n-1}\)

\(\Rightarrow\) (i) 8th term = \(ar^{8-1} = ar^7\),

(ii) 15th term = \(ar^{14}\),

(iii) 20th term = \(ar^{19}\) and so on.

If out of the four quantities a, r, n and \(t_n\) (i.e. l) any three quantities are given, we can find the remaining fourth quantity.

Problem Set 1

Find which of the following is a G.P.:

(i) \(2, 2\sqrt{2}, 4, 4\sqrt{2}, \ldots\)

(ii) \(\frac{1}{3}, \frac{2}{3}, 1, \frac{4}{3}, \ldots\)

(iii) 4, 8, 16, ...

(iv) xy, x²y, x³y, ...

Solution

(i) Since, \(\frac{2\sqrt{2}}{2} = \sqrt{2}, \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}, \frac{4\sqrt{2}}{4} = \sqrt{2}\)

\(\Rightarrow \frac{2\sqrt{2}}{2} = \frac{4}{2\sqrt{2}} = \frac{4\sqrt{2}}{4} = \ldots = \sqrt{2}\)

\(\therefore 2, 2\sqrt{2}, 4, 4\sqrt{2}, \ldots\) is a G.P. with common ratio \(\sqrt{2}\). Ans.

(ii) Since, \(\frac{2/3}{1/3} \neq \frac{1}{2}\), given sequence is not a G.P.

(iii) Since, \(\frac{8}{4} = \frac{16}{8} = \ldots = 2\),

\(\therefore\) Sequence 4, 8, 16, ... forms a G.P. with common ratio 2. Ans.

(iv) It is a G.P. as

\(\frac{x^2y}{xy} = \frac{x^3y}{x^2y} = \ldots = x\). Ans.

Problem 2

Find the 8th term of the geometric progression: 5, 10, 20, ... .

Solution

Clearly, first term (a) = 5

and, common ratio (r) = \(\frac{10}{5}\) = 2

\(\therefore t_n = ar^{n-1} \Rightarrow t_8 = 5 \times 2^{8-1}\)

= \(5 \times 2^7 = 5 \times 128 =\) 640 Ans.

Problem 3

Find the 19th term of the series: \(\sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \ldots\)

Solution

In the given series

\(\frac{t_2}{t_1} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{3}, \frac{t_3}{t_2} = \frac{\frac{1}{3\sqrt{3}}}{\frac{1}{\sqrt{3}}} = \frac{1}{3}\) and so on.

\(\Rightarrow \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \ldots\) is a G.P. with common ratio \(\frac{1}{3}\).

Since, first term a = \(\sqrt{3}\), common ratio r = \(\frac{1}{3}\) and n = 19

\(\therefore\) 19th term = \(ar^{18}\)

[∴ nth term = \(ar^{n-1}\)]

= \(\sqrt{3} \times \left(\frac{1}{3}\right)^{18} = \frac{\sqrt{3}}{3^{18}}\) Ans.

Problem 4

If the first two consecutive terms of a G.P. are 125 and 25, find its 6th term.

Solution

Clearly, first term a = 125 and second term ar = 25

\(\therefore\) Common ratio = \(\frac{ar}{a} = \frac{25}{125} \Rightarrow r = \frac{1}{5}\)

\(\therefore\) 6th term of the G.P. = \(ar^5\)

= \(125 \times \left(\frac{1}{5}\right)^5 = \frac{125}{3125} = \frac{1}{25}\) Ans.

Problem 5

Find the next three terms of the sequence: 36, 12, 4, ... .

Solution

Since, \(\frac{12}{36} = \frac{1}{3}, \frac{4}{12} = \frac{1}{3}, \ldots\)

i.e., \(\frac{12}{36} = \frac{4}{12} \Rightarrow 36, 12, 4, \ldots\) form a G.P.

Clearly, first term a = 36 and common ratio r = \(\frac{1}{3}\)

\(\therefore\) Next three terms = 4th term, 5th term and 6th term

= \(ar^3, ar^4\) and \(ar^5\)

= \(36 \times \left(\frac{1}{3}\right)^3, 36 \times \left(\frac{1}{3}\right)^4\) and \(36 \times \left(\frac{1}{3}\right)^5\)

= \(\frac{4}{3}, \frac{4}{9}\) and \(\frac{4}{27}\) Ans.

Teacher's Note

Geometric progressions appear in real life when quantities grow or shrink by a constant factor, such as bacterial growth, radioactive decay, or investment returns compounded regularly.

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