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Geometric Progression


Geometric Progression is the sequence of numbers such that the next term of the sequence comes by multiplying or dividing the preceding number with the constant (non-zero) number. And the constant number is called the Common Ratio. It is also known as the Geometric Sequence.

geometric progression

a, ar, ar2, ar3,……..,arn

The first term of the sequence is called the Initial Term or the scale factor which is denoted as ‘a’.

The ratio of a term to its next term of the sequence is called the Common ratio which is denoted by ‘r’.

The general term that is the nth term of the geometric progression with the initial term ‘a’ and the common ratio ‘r’ is $ \displaystyle {{a}_{n}}=a\cdot {{r}^{{n-1}}}$.

Example 1: Find the nth term of 3, 6, 12, 24, 48,……….aif the number of terms are 10?

Solution: Given the progression  3, 6, 12, 24, 48,……….aand n=10.
By the general formula of the geometric progression $ \displaystyle {{a}_{n}}=a\cdot {{r}^{{n-1}}}$

The first term a=3
Common ratio is r=6/3=2
Therefore $ \displaystyle {{a}_{n}}=3\cdot {{2}^{{10-1}}}=3\cdot {{2}^{9}}=3\cdot 512=1536$

Finite and Infinite Geometric Progression

The Geometric Progression with a limited number of terms is called Finite Sequence. It has a last term.

Example: 1, 3, 9, 27,…..729

This is a finite sequence with a=1 and r=3

The Geometric Progression with unlimited number of terms is called Infinite Sequence. It does not have a last term.

Example: 1, 5, 25, 125,……..
This is an infinite sequence with a=1 and r=5

How to find the common ratio?

The common ratio is the ratio of the term of the geometric sequence to its previous term.And is denoted by “r”.
To find the common ratio we need to take the ratio of the terms with their preceding term.
$ \displaystyle \frac{{{{a}_{2}}}}{{{{a}_{1}}}}=\frac{{{{a}_{3}}}}{{{{a}_{2}}}}=………..\frac{{{{a}_{n}}}}{{{{a}_{{n-1}}}}}$

Example 2: What is the common ratio of the sequence 1, 2, 4, 8,……..?

Solution: We have
a1=1,  a2=2,  a3=4,  a4=8

$ \displaystyle r=\frac{{{{a}_{2}}}}{{{{a}_{1}}}}=\frac{{{{a}_{3}}}}{{{{a}_{2}}}}=\frac{{{{a}_{4}}}}{{{{a}_{3}}}}$

$ \displaystyle r=\frac{2}{1}=\frac{4}{2}=\frac{8}{4}=2$

Therefore the common ratio is 2.

Tip! When we want to check if the sequence is Geometric or not we can use the formula of common ratio.

Example 3: Check whether the sequence$ \displaystyle \frac{1}{3},\frac{1}{6},\frac{1}{{12}},\frac{1}{{24}},……..$ is geometric progression or not?

Solution: We have:

$ \displaystyle {{a}_{1}}=\frac{1}{3}$

$ \displaystyle {{a}_{2}}=\frac{1}{6}$

$ \displaystyle {{a}_{3}}=\frac{1}{{12}}$

$ \displaystyle {{a}_{4}}=\frac{1}{{24}}$

Let’s find the common ratio of our sequence.

$ \displaystyle r=\frac{{{{a}_{2}}}}{{{{a}_{1}}}}=\frac{1}{6}\div \frac{1}{3}=\frac{1}{2}$

$ \displaystyle r=\frac{{{{a}_{3}}}}{{{{a}_{2}}}}=\frac{1}{{12}}\div \frac{1}{6}=\frac{1}{2}$

$\displaystyle r=\frac{{{{a}_{4}}}}{{{{a}_{3}}}}=\frac{1}{{24}}\div \frac{1}{{12}}=\frac{1}{2}$

Since the ratio of all the terms with their preceding terms is constant $\displaystyle \frac{1}{2}$ then this is a Geometric Progression.

Formula of the nth term 

If a is the first term, r is the common ratio and n is the total number of the terms, then the formula for the nth term is $ \displaystyle {{a}_{n}}=a\cdot {{r}^{{n-1}}}$

Example 4: Find the 7th term of the geometric progression 2,4,8,16…

Solution: We have a=2 and r=2
All we have to do is substitute the values in the formula of the nth term.
$ \displaystyle {{a}_{n}}=a\cdot {{r}^{{n-1}}}$
$ \displaystyle {{a}_{7}}=2{{\left( 2 \right)}^{{7-1}}}=2\cdot {{2}^{6}}=2\cdot 64=128$
So the 7th term of the geometric progression is 128.

Sum of a Finite Geometric Progression

IF “a” is the first term and “r”  is the common ratio of the sequence with “n” number of terms where r≠1 than the sum of the sequence is   a + ar + ar+ ar+…………+ arn

The formula of the sum: $ \displaystyle {{s}_{n}}=\frac{{{{a}_{1}}(1-{{r}^{n}})}}{{1-r}}$

Example 5: Find the sum of the given geometric progression 1, 2, 4, 8, 16, 32 ?

Solution: We have 6 number of terms in total so n=6
The first term is a1=1 and the common ratio is r=2
All we have to do is substitute the values in the formula
$ \displaystyle {{s}_{n}}=\frac{{{{a}_{1}}(1-{{r}^{n}})}}{{1-r}}$
$ \displaystyle {{s}_{n}}=\frac{{1(1-{{2}^{6}})}}{{1-2}}$
$ \displaystyle {{s}_{n}}=\frac{{1-64}}{{-1}}$
$ \displaystyle {{s}_{n}}=\frac{{-63}}{{-1}}$
$ \displaystyle {{s}_{n}}=63$
So the sum is 63.

Tip! If r=1 then the sequence will be in the form a, a, a, a, a,…..a and the formula of the sum will be Sn=a × n

Sum of an Infinite Geometric Progression

If we have to find the sum of the infinite geometric progression then we have a different formula.
When -1<r<1 and n is infinite that is ∞ then rn will tend to zero
 So our formula transforms to $ \displaystyle {{s}_{\infty }}=\frac{{{{a}_{1}}}}{{1-r}}$

Example 6: Find the sum of the geometric progression  $\displaystyle 1,\frac{1}{2},\frac{1}{4},\frac{1}{8}………….$ ?

Solution: We have $ \displaystyle {{a}_{1}}=1$ and  $ \displaystyle r=\frac{1}{2}$
Since -1<r<1 then we use the formula $ \displaystyle {{s}_{\infty }}=\frac{{{{a}_{1}}}}{{1-r}}$
All we have to do is substitute the values in the formula
$ \displaystyle {{s}_{\infty }}=\frac{1}{{1-\frac{1}{2}}}=\frac{1}{{\frac{1}{2}}}=2$
So the sum is 2.

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