teamryan15 wrote:
Bunuel wrote:
In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5-a_2=12\), what is the value of a_1?
A. 4
B. 24/7
C. 2
D. 12/7
E. 6/7
The formula for calculating \(n_{th}\) term would be \(a_n=2^{n-1}*a_1\) . So:
\(a_5=2^4*a_1\);
\(a_2=2*a_1\);
Given: \(a_5-a_2=2^4*a_1-2*a_1=12\) --> \(2^4*a_1-2*a_1=12\) --> \(a_1=\frac{12}{14}=\frac{6}{7}\).
Answer: E.
Hope it's clear.
Can someone help me out here? I dont know if I am reading the question correctly but it says each term after the first term is equal to
twice the previous term but the solution above shows it as two times the previous term times the first term. I am not seeing how that makes sense?
Shouldnt it be \(a_5=2^4?\)
teamryan15 I think I see where you're a little off. The first term is not 1. You are focused on just the coefficient / multiplier, I think.
If \(A_1\)
were 1, then yes, \(A_5\) would = \(2^4\). We would have:
\(A_1 = 1\)
\(A_2 = 2\)
\(A_3 = 4\)
\(A_4 = 8\)
\(A_5 = 16 = 2^4\)
The first term is \(a\), not 1. Thus:
\(A_1 = a_1\)
\(A_2 = (2*a_1) = 2a_1 = 2^1*a_1\)
\(A_3 = (2*2a_1)= 4a_1 = 2^2*a_1\)
\(A_4 = (2*4a_1)= 8a_1 = 2^3* a_1\)
\(A_5 = (2*8a_1) = 16a_1 = 2^4*a_1\)
Hope that helps.
_________________
—The only thing more dangerous than ignorance is arrogance. ~Einstein—I stand with Ukraine.
Donate to Help Ukraine!