 # How To Solve Geometric Progression

## Introduction of Geometric Progression

If a sequence of terms is such that each term is constant multiple of the preceding term, then the sequence is called a geometric progression (G.P.).

The constant multiplier is called the common ratio.

### Examples of Geometric Progression

5, 15, 45, 135, common ratio 3. 1, ½, ¼, 1/8, … common ratio ½. 2, -6, 18, -54, … common ratio -3.

## General Form of a Geometric Progression

Consider the sequence a, ar, ar2, ar3,…. First term = a, Second term = ar = ar2 – 1, Third term = ar2 = ar3 – 1, Fourth term = ar3 = ar4 – 1. Similarly nth term = arn – 1.

$\therefore {{t}_{n}}=a{{r}^{n-1}}$

## Geometric Mean

If x, y, z are three consecutive terms of geometric progression, then common ratio

$r=\frac{y}{x}=\frac{z}{y}$

$\Rightarrow {{y}^{2}}=xz$

$\therefore y=\sqrt{xz}$

This y is called the geometric mean of x and z.

## Sum of n terms of a Geometric Progression

Let Sn denote the sum of n terms of a G.P. whose first term = a and common ratio = r

Sn = a + ar + ar2 + … + arn – 2 + arn – 1 ……….(1)

r. Sn =  ar + ar2 + … + arn – 1 + arn ……….(2) [Multiplying (1) by r]

By subtracting (2) from (1) we get,

${{S}_{n}}-{{S}_{n}}.r=a-a{{r}^{n}}$

$\Rightarrow {{S}_{n}}\left( 1-r \right)=a\left( 1-{{r}^{n}} \right)$

$\therefore {{S}_{n}}=\frac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$

if r < 1, both numerator and denominator of this fraction are positive and this is the most convenient form of Sn. If however, r > 1, both numerator and denominator are negative and more convenient form is ${{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}$

## Three or More Terms in G.P.

Three numbers are in G.P. as

$\frac{a}{r},a,ar$

Four numbers are in G.P. as

$\frac{a}{{{r}^{3}}},\frac{a}{r},ar,a{{r}^{3}}$

Five numbers are in G.P. as

$\frac{a}{{{r}^{2}}},\frac{a}{r},a,ar,a{{r}^{2}}$

## Properties of G.P. Series

If a1, a2, …, an be in G.P. and k be a constant quantity (≠ 0), then

$(i)~~{{a}_{1}}k,{{a}_{2}}k,{{a}_{3}}k,…~~\text{are also in G}.\text{P}.$

$(ii)~~\frac{{{a}_{1}}}{k},\frac{{{a}_{2}}}{k},…,\frac{{{a}_{n}}}{k}~~\text{are also in G}.\text{P}.$

$(iii)~~\frac{1}{{{a}_{1}}},\frac{1}{{{a}_{2}}},…,\frac{1}{{{a}_{n}}}~~\text{are also in G}.\text{P}.$

$(iv)~~a_{1}^{m},a_{2}^{m},…,a_{n}^{m}~~\text{are also in G}.\text{P}.$

 Example 01

Find the 12th term and the sum of first 12 terms of the G.P. 3, 6, 12, 24, …

Solution:

Here

$a=3,~~r=\frac{6}{3}=2,~~nth~~term=a{{r}^{n-1}}$

$\therefore 12th~~term=3{{\left( 2 \right)}^{12-1}}=3\times {{2}^{11}}=6144$

Now,

${{S}_{12}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}=\frac{3\left( {{2}^{12}}-1 \right)}{2-1}=3\left( 4096-1 \right)=12285$

 Example 02

Find the sum of the series 4 + 44 + 444 + 4444 + … to n terms.

Solution:

$Let~~{{S}_{n}}=4+44+444+4444+…~~to~~n~~terms.$

$\Rightarrow {{S}_{n}}=4\left[ 1+11+111+1111+…~~to~~n~~terms \right]$

$\Rightarrow {{S}_{n}}=\frac{4}{9}\left[ 9+99+999+9999+…~~to~~n~~terms \right]$

$=\frac{4}{9}\left[ \left( 10-1 \right)+\left( 100-1 \right)+\left( 1000-1 \right)+…~~to~~n~~terms \right]$

$=\frac{4}{9}\left[ \left( 10+100+1000+…~~to~~n~~terms \right)-n \right]$

$=\frac{4}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right]$

$\therefore {{S}_{n}}=\frac{40}{81}\left( {{10}^{n}}-1 \right)-\frac{4n}{9}$

 Example 03

If the 1st, 2nd and 5th terms of an A.P. are three consecutive terms of a G.P., find the common ratio.

Solution:

Let the terms of the G.P. be a, ar and ar2. If d is the common difference of the A.P. then

$ar-a=d……….(i)$

$a{{r}^{2}}-a=4d……..(ii)$

Using (i) and (ii) we get,

$a{{r}^{2}}-a=4\left( ar-a \right)$

$\Rightarrow {{r}^{2}}-1=4r-4~~\left[ \because a\ne 0 \right]$

$\Rightarrow {{r}^{2}}-4r+3=0$

$\Rightarrow \left( r-1 \right)\left( r-3 \right)=0$

$\therefore r=1~~or~~3$

If r = 1, the three terms are identical. Hence, common ratio = 3.

 Example 04

If S be the sum P be the product and R the sum of reciprocals of n terms in G.P., prove that

${{\left( \frac{S}{R} \right)}^{n}}={{P}^{2}}$

Solution:

Let a be the first term and r the common ratio of the G.P.

Now,

$S=a+ar+a{{r}^{2}}+…+a{{r}^{n-1}}$

$\Rightarrow S=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}……….(i)$

$P=a\times ar\times a{{r}^{2}}\times …\times a{{r}^{n-1}}$

$\Rightarrow P={{a}^{n}}\times {{r}^{1+2+3+…+\left( n-1 \right)}}$

$\Rightarrow P={{a}^{n}}\times {{r}^{\frac{n\left( n-1 \right)}{2}}}……….(ii)$

And

$R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{{r}^{2}}}+…+\frac{1}{a{{r}^{n-1}}}$

$\Rightarrow R=\frac{\frac{1}{a}\left\{ 1-{{\left( \frac{1}{r} \right)}^{n}} \right\}}{1-\frac{1}{r}}=\frac{\frac{1}{a}\left\{ 1-\frac{1}{{{r}^{n}}} \right\}}{\frac{r-1}{r}}$

$\Rightarrow R=\frac{1}{a}\times \frac{{{r}^{n}}-1}{{{r}^{n}}}\times \frac{r}{r-1}=\frac{r\left( {{r}^{n}}-1 \right)}{a{{r}^{n}}\left( r-1 \right)}$

$\Rightarrow R=\frac{{{r}^{n}}-1}{a{{r}^{n-1}}\left( r-1 \right)}……….(iii)$

$\therefore \frac{S}{R}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\times \frac{a{{r}^{n-1}}\left( r-1 \right)}{{{r}^{n}}-1}={{a}^{2}}.{{r}^{n-1}}$

$\therefore {{\left( \frac{S}{R} \right)}^{n}}={{a}^{2n}}.{{r}^{n\left( n-1 \right)}}$

And

${{P}^{2}}={{\left[ {{a}^{n}}\times {{r}^{\frac{n\left( n-1 \right)}{2}}} \right]}^{2}}={{a}^{2n}}.{{r}^{n\left( n-1 \right)}}$

$\therefore {{\left( \frac{S}{R} \right)}^{n}}={{P}^{2}}$

 Example 05

Find three numbers in G.P. whose sum is 13 and the sum of whose squares is 91.

Solution:

Let the required numbers be a/r, a and ar.

Then,

$\frac{a}{r}+a+ar=13……….(i)$

$\frac{{{a}^{2}}}{{{r}^{2}}}+{{a}^{2}}+{{a}^{2}}{{r}^{2}}=91……….(ii)$

On squaring both sides of (i) we get,

${{\left( \frac{a}{r}+a+ar \right)}^{2}}=169$

$\Rightarrow {{\left( \frac{a}{r} \right)}^{2}}+{{\left( a \right)}^{2}}+{{\left( ar \right)}^{2}}+2\left\{ \left( \frac{a}{r} \right).a+\left( \frac{a}{r} \right).\left( ar \right)+a\left( ar \right) \right\}=169$

$\Rightarrow \frac{{{a}^{2}}}{{{r}^{2}}}+{{a}^{2}}+{{a}^{2}}{{r}^{2}}+2\left\{ \frac{{{a}^{2}}}{r}+{{a}^{2}}+{{a}^{2}}r \right\}=169$

$\Rightarrow 91+2a\left\{ \frac{a}{r}+a+ar \right\}=169$

$\Rightarrow 91+2a\times 13=169$

$\Rightarrow 26a=78$

$\therefore a=3$

Putting a = 3 in (i) we get,

$\frac{3}{r}+3+3r=13$

$\Rightarrow \frac{3+3r+3{{r}^{2}}}{r}=13$

$\Rightarrow 3{{r}^{2}}+3r+3=13r$

$\Rightarrow 3{{r}^{2}}-10r+3=0$

$\Rightarrow 3{{r}^{2}}-9r-r+3=0$

$\Rightarrow 3r\left( r-3 \right)-1\left( r-3 \right)=0$

$\Rightarrow \left( r-3 \right)\left( 3r-1 \right)=0$

$\therefore r=3~~or~~\frac{1}{3}$

Therefore, the three numbers are 1, 3, 9 or 9, 3, 1.

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### 2 thoughts on “How To Solve Geometric Progression”

1. Thanks nice post. I will be happy if you write post on harmonic progression. 😊

1. Thanks for your comment, I will definitely write a post on Harmonic progression ;). 0

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