ALGEBRA Geometric Progression

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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