# General term of Geometric Progression

The general term of geometric progression with first term 'a' and the common ratio is 'r' is given by
$a_{n} = a r^{n -1}$
Where 'n' is a number of terms.
Theorem : Prove that the nth term of G.P. with first term 'a' and the common ratio 'r' is given by $a_{n} = a r^{n -1}$
Proof: Let $a_{1},a_{2},a_{3},a_{4},...,a_{n}$ be the given G.P. then
$a_{1}$ = a ⇒ a $r^{1 -1}$ = a$r^{0}$ = a
As 'r' is a common ratio
∴ $\frac{a_{2}}{a_{1}}$ = r ⇒ $a_{2} = a_{1}$r⇒ $a_{2}$ = ar

$\frac{a_{3}}{a_{2}} = r ⇒ a_{3} = a_{2} r ⇒ a_{3} = (ar)r = ar^{2}=ar^{3-1}$

$\frac{a_{4}}{a_{3}} = r ⇒ a_{4} = a_{3} r⇒ a_{4} = (ar^{2})r = ar^{3} = ar^{4-1}$
Continuing in this manner we get ,
$a_{n} = ar^{n - 1}$
So the sequence will be a, ar, a$r^{2}, ,....,ar^{n-1}$ according as it is finite or infinite.

## Examples on general term of geometric progression

Example 1 : Write the general term of geometric progression if the sequence is 1,4,16,64,...
Solution : General term is given by
$a_{n} = ar^{n - 1}$ ------------(1)
According to the given sequence
first term = a = 1 ; common ratio = r = 4
∴ equation (1) ⇒ $a_{n} = 1 \times 4^{n -1}$
$a_{n} = 4^{n -1}$

Example 2 : Write the general term of geometric progression if the sequence is $\frac{1}{4} , \frac{-1}{2}$,1, -2, ...

Solution : General term is given by
$a_{n} = ar^{n - 1}$ ------------(1)
According to the given sequence
first term = a = $\frac{1}{4}$ ; common ratio = r = -2
∴ equation (1) ⇒ $a_{n} = \frac{1}{4} \times (-2)^{n -1}$

$a_{n} = 2^{-2} \times (-2)^{n -1}$

$a_{n} = 2^{-2} \times 2^{n-1} \times(-1)^{n-1}$

$a_{n} = (-1)^{n -1} 2^{n -3}$
Example 3 : Write the general term of geometric progression if the sequence is 3,6,12,24, ...

Solution : General term is given by
$a_{n} = ar^{n - 1}$ ------------(1)
According to the given sequence
first term = a = 3 ; common ratio = r = 2
∴ equation (1) ⇒ $a_{n} =3 \times 2^{n -1}$
$a_{n} = 3 \times 2^{n -1}$