we have an geometric progression:
nth term=a₁r^(n-1)
nth term=y
a₁ is the first term=10
r=common ratio=a₂/a₁=20/10=2
n=x
nth term=a₁r^(n-1)
Therefore
y=10.2^(x-1)=(2*5)2^(x-1)=(5)2^(x-1+1)=(5)2^x
Answer: y=[5][(2)]^[x]
To check:
if x=1 ⇒y=(5)(2)=10
if x=2 ⇒y=(5)(4)=20
if x=3 ⇒y=(5)(8)=40