Hello BIGChungus! I like your name!
Plan:
For arithmetic problems, I like using variables.
First term = x
Common difference = y
We can define an arithmetic sequence this way:
x + (x + y) + (x + 2y) + (x + 3y)....
Set up and solve:
(1) \(\frac{a_4}{a_2}=\frac{x+3y}{x+y}=3\)
(2) \(\frac{a_5}{a_3}=\frac{x+4y}{x+2y}=?\)
Let us get rid of the fraction in (1) by multiplying both sides by (x+y)
x + 3y = 3x + 3y
-2x = 0
x = 0
Wow! The first term is 0.
Let us plug that into (2).
\(\frac{0+4y}{0+2y}\)
Simplify
\(\frac{4y}{2y}=2\)
Ta-da! Mathz!