Assume the first two numbers are A and A
The row will be
A, A, 2A, 4A, 8A
For the nth number in the list it can be represented in the way number=[tex]A*(2)^{n-2}[/tex] for any n≥3
Using this formula the answer of the question can be obtained with division of the 23th and 15th term
[tex]\frac{A*(2)^{23-2}}{A*(2)^{15-2}}[/tex]
Simplify
=[tex]\frac{A*(2)^{21}}{A*(2)^{13}}[/tex]
=[tex]\frac{2^{21}}{2^{13}}[/tex]
=[tex]2^{21-13}[/tex]
=[tex]2^8[/tex]
=256
Done!
