In mathematics, the geometric progression is a sequence of numbers in which each term is driven from the preceding term by multiplying a fixed number, where the fixed number which is multiplied with each term is called "common difference" and it is denoted by "d". The common difference can not be zero.
We can get a common difference (d) by dividing any previous term to the next term of any GP series.
Example(1): 2, 4, 8, 16, 32, ......
Here 2 is multiplied with each preceding term to get the next term so 2 is a common difference (d).
Example(2): 3, 9, 27, 81, 243, ......
Here 3 is multiplied with each preceding term to get the next term so 3 is a common difference (d).
Let's consider the first term of a series is "a" and the common difference is "d" then the general term of geometric progression will be $$ a, \ a \times d, \ a \times d^2, \ a \times d^3, \ a \times d^4..... $$ It can be also written as
\(a \times d^0, \ a \times d^1, \ a \times d^2, \ a \times d^3, \) \(a \times d^4....\)
Further this series can be written as
\(a \times d^{1-1}, \ a \times d^{2-1}, \ a \times d^{3-1}, \) \(a \times d^{4-1}, \ a \times d^{5-1}....\)
Hence, \(n^{th}\) term of the series $$ T_n = a \times d^{n-1} $$
Example(3): Find the \(7^{th}\) term of the given GP series? $$ 2, 4, 8, 16,...... $$
Solution: From the given series the first term of the series (a) = 2
The common difference (d) = \(\frac{4}{2}\) = 2
Then, $$ T_n = a \times d^{n-1} $$ $$ T_7 = 2 \times 2^{7-1} $$ $$ T_7 = 2 \times 2^6 $$ $$ T_7 = 2 \times 64 $$ $$ T_7 = 128 $$ Hence the \(7^{th}\) term of the series will be 128.
Example(4): Find the \(8^{th}\) term of the given GP series? $$ 1, 2, 4, 8, 16,...... $$
Solution: From the given series the first term of the series (a) = 1
The common difference (d) = \(\frac{2}{1}\) = 2
Then, $$ T_n = a \times d^{n-1} $$ $$ T_8 = 1 \times 2^{8-1} $$ $$ T_8 = 1 \times 2^7 $$ $$ T_8 = 1 \times 128 $$ $$ T_8 = 128 $$ Hence the \(8^{th}\) term of the series will be 128.