ov25 wrote:For the geometric sequence, a1, a1*d, a1*d^2, ... a1*d^n, the sum of the first three terms is
21, what is the sum of the first 6 terms?
(1) a1=3
(2) d>0

discuss

a1*d(1+d)+a1=21, to find the sum of the first six we need to find d
st(1) a1=3 <=> 3*d(1+d) +3=21 <=> 3d+3*d^2=18 <=> d^2+d-6=0 <=> (d+3)(d-2)=0 => 2 answers for d {-3:2} Not sufficient
st(2) d>0. Not sufficient.

Combining st(1&2) d=2 Sufficient.

IOM C

Answer: C (both together are sufficient).

as per given condition: a + ad + ad^2 = 21 (=S3)
a*(1 + d + d^2) = 21
Hence To calculate the sum of 6 terms, we need to know both a & d.
lets check the given conditions:

I: a = 3: it means: (1+ d+ d^2) = 7
solve to get: (d+3)(d-2) = 0 -> d = -3, 2 NOT SUFFICIENT.

II: d > 0: NOT SUFFICENT (value of a is still wide open)

Take I & II together: d = 2, a = 3 SUFFICIENT. Hence (D)

Let me know if you have questions.
Regards,
Bharat.