QUESTION 10
Easy
Which term of the GP: 2, 8, 32, ... is 131072? Write the explicit formula as well as the recursive formula for the \(n^{th}\) term.
SOLUTION
1
Identify the values in the GP
The GP is: 2, 8, 32,...
Here, \(a = 2,\,r = \frac{8}{2} = 4\)
Here, \(a = 2,\,r = \frac{8}{2} = 4\)
2
Find \(n\) using the \(n^{th}\) term formula
\(T_n = ar^{n−1}\)
⇒ \(131072 = 2(4)^{n−1}\)
⇒ \(65536 = 4^{n−1}\)
Since \(65536 = 2^{16}\) and \(4^{n−1} = (2^2)^{n−1} = 2^{2(n−1)}\)
\(2^{16} = 2^{2(n−1)}\)
⇒ \(2(n − 1) = 16\)
⇒ \(n = 9\)
So, 131072 is the \(9^{th}\) term.
Explicit formula:
\(T_n = 2(4)^{n−1}\)
Recursive formula:
\(T_1 = 2,\,T_2 = 4T_{n−1}\)
⇒ \(131072 = 2(4)^{n−1}\)
⇒ \(65536 = 4^{n−1}\)
Since \(65536 = 2^{16}\) and \(4^{n−1} = (2^2)^{n−1} = 2^{2(n−1)}\)
\(2^{16} = 2^{2(n−1)}\)
⇒ \(2(n − 1) = 16\)
⇒ \(n = 9\)
So, 131072 is the \(9^{th}\) term.
Explicit formula:
\(T_n = 2(4)^{n−1}\)
Recursive formula:
\(T_1 = 2,\,T_2 = 4T_{n−1}\)
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Final Answer :
131072 is the \(9^{th}\) term.
Explicit formula: \(T_n = 2(4)^{n−1}\)
Recursive formula: \(T_1 = 2,\,T_2 = 4T_{n−1}\)
Concept Note
For a GP,
\(T_n = ar^{n−1}\)
To find the position of a term, equate \(T_n\) to the given value and solve for \(n\).