QUESTION 6
Easy
Which term of the sequence 2, 2√2, 4, ... is 128?
SOLUTION
1
Write down the values
In the GP: 2, 2√2, 4, ...
\(a = 2\)
\(r = \frac{2√2}{2} = √2\)
\(a = 2\)
\(r = \frac{2√2}{2} = √2\)
2
Use GP formula
\(128 = 2(2√2)^{n−1}\)
⇒ \(64 = (√2)^{n−1}\)
We know,
\(64 = 2^6\) and \(√2 = 2^{\frac{1}{2}}\)
So, \( (2^{\frac{1}{2}})^{n−1} = 2^6\)
⇒ \(2^{\frac{n−1}{2}} = 2^6\)
Equating the powers, we get
\(\frac{n−1}{2} = 6\) ⇒ \(n − 1 = 12\)
⇒ \(n = 13\)
So, 128 is the \(13^{th}\) term.
⇒ \(64 = (√2)^{n−1}\)
We know,
\(64 = 2^6\) and \(√2 = 2^{\frac{1}{2}}\)
So, \( (2^{\frac{1}{2}})^{n−1} = 2^6\)
⇒ \(2^{\frac{n−1}{2}} = 2^6\)
Equating the powers, we get
\(\frac{n−1}{2} = 6\) ⇒ \(n − 1 = 12\)
⇒ \(n = 13\)
So, 128 is the \(13^{th}\) term.
🏆
Final Answer : 128 is the \(13^{th}\) term of the GP.
Concept Note
When roots appear in a GP, rewrite them as powers.
\(√2 = 2^{\frac{1}{2}}\)
Then compare the exponents to solve for \(n\).