It has to be a or b since putting a 1 in for n gives
[tex]2*5^(^1^-^1^)=2*5^0=2*1=2[/tex]
If you put n-1 in for n you get
[tex]a_{n-1}=2*5^{(n-1-1)}=2*5^{(n-2)}[/tex]
Now for any a_n we need the total sequence before it as well as the nth value. But the total sequence before it is just a_(n-1):
[tex]2*5^{(n-2)}*5=2*5^{(n-1)}[/tex]
so we need choice a, since the (n-1)th term times 5 is the nth term
