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


\[\Rightarrow {{y}^{2}}=xz\]

\[\therefore y=\sqrt{xz}\]

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,


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

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

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


Four numbers are in G.P. as


Five numbers are in G.P. as



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, …




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


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

Example 02

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



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

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

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

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

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

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

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.


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



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

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


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



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

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

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

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



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

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

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

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

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


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

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

Example 05

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


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




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,


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

\[\Rightarrow 3{{r}^{2}}+3r+3=13r\]

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

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

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

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

\[\therefore r=3~~or~~\frac{1}{3}\]

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

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