QUESTION 4
Easy
Which term of the GP: 2, 6, 18,... is 4374? Write the explicit formula as well as the recursive formula for the \(n^{th}\) term.
SOLUTION
1
Identify the values
In the GP: 2, 6, 18,...
first term, \(a = 2\)
common ratio, \(r = \frac{6}{2} = 3\)
first term, \(a = 2\)
common ratio, \(r = \frac{6}{2} = 3\)
2
Use the GP formula to find out which term is 4374
\(4374 = 2(3)^{n − 1}\)
⇒ \(2187 = 3^{n − 1}\)
Since \(2187 = 3^7\),
\(n − 1 = 7\)
⇒ \(n = 8\)
So, 4374 is the \(8^{th}\) term.
⇒ \(2187 = 3^{n − 1}\)
Since \(2187 = 3^7\),
\(n − 1 = 7\)
⇒ \(n = 8\)
So, 4374 is the \(8^{th}\) term.
3
Write the explicit formula
\(T_n = 2(3)^{n − 1}\)
4
Write the recursive formula
\(T_1 = 2\)
\(T_{n + 1} = 3T_n\)
\(T_{n + 1} = 3T_n\)
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Final Answer :
4374 is the \(8^{th}\) term.
Explicit formula: \(T_n = 2(3)^{n − 1}\)
Recursice formula: \(T_1 = 2\)
\(T_{n + 1} = 3T_n\)
Concept Note
For a GP,
\(T_n = ar^{n − 1}\)
To find the position of a term, set \(T_n\) equal to the given value and then solve for \(n\).