D IMO

(1) Each term after the first has a (4*5) factor. Hence all terms after the 1st would be divisible by 20.
(2) a4,a5,a6 are given. We can find the common multiple and all the terms in GP series itself. Sufficient. (all terms have a factor of 5*4 after the first)

THis problem does not appear GMAT like for the fact, (1) mentions a1=5 but using (2) we calculate a1=4. Usually don't see such contradictions in the GMAT.

I dont think we can assume a1 as 4, IMO:A

agree with Shankar, it's not GMAT like q. as we don't need to spend time to find function for sequence in st(2); we only supply the previous st(1) function and see it's correctly working here
also agree with gmatblood, we can't assign a(1)=4 as the function isn't clear if n is integer or ...
weird DS, what's the source

st(1) a(n)=4* 5^(n-1) contains 4 and 5, i.e. factors of 20 Sufficient
st(2) Sufficient!?

d

GmatKiss wrote:Does the sequence {a1, a2, a3, ..., an, ...} contain an infinite number of terms that are divisible by 20?

(1) a1 = 5 and an = 4(5^(n - 1)) for all integers n ≥ 2.

(2) a2 = 20, a4 = 500, a5 = 2,500, and a6 = 12,500.

From (2) a1 cannot take any value other than 4 here.

You know 3 consecutive terms of the series are (500,2500,12500)
Each term is clearly multiplied by 5. Don't think any other mathematical operation would produce this series. If a1 is not 4, we would never have a2=20,a4=500. The series would not hold good at all for any other value of a1.

A general GP would be a,ar,arr,arrr ...

you know r=5; and ar=20. a should be 4

I think it is A.

1) As a1 is 5, it is not divisible by 20. So we can say that all numbers of the set are not divisible by 20.
2) The Question does not say the numbers are in any progression. So giving a few samples does not help

Regards,