The answer is 1, -2, 4, -8, 16, or the same terms in reverse order.

Proof is as follows: take the first n terms of the sequence (we will solve for n later). Then we have:

a(1 + r + r^{2} + ... + r^{n}-1) = 11

a^{2}(1 + r^{2} + r^{4} + ... + r^{2n-2}) = 341

a^{3}(1 + r^{3} + r^{6} + ... + r^{3n-3}) = 3641

Using the standard geometric progression formulae,

a (r^{n} - 1)/(r-1) = 11

a^{2} (r^{2n} - 1)/(r^{2}-1) = 341

a^{3} (r^{3n} - 1)/(r^{3}-1) = 3641

We want to get rid of the a's, and the simplest way to do this is to divide the second equation by the first one

squared:

(r^{2n} - 1) (r - 1)^{2} / ( (r^{2} - 1)(r^{n} -

1)^{2}) =

341/112 =

31/11

Now we have an equation with only two unknowns, r and n. I'm going to put u = r^{n} and solve for it in terms of r.

We have:

(u^{2} - 1) (r - 1)^{2} / ( (r^{2} - 1)(u -

1)^{2}) = 31/11

(u+1)(u-1)(r-1)^{2} / ( (r+1)(r-1)(u-1)^{2}) = 31/11

(u+1)(r-1) / ( (r+1)(u-1) ) = 31/11

11(u+1)(r-1) = 31(r+1)(u-1)

11ur-11u+11r-11 = 31ur-31r+31u-31

20ur+42u = 42r+20

u(20r+42) = 42r+20

u = (42r+20)/(20r+42)

u = (21r+10)/(10r+21)

Now, let's repeat the whole process. Go back to the three equations we started with and divide the third by the cube

of the first.

We have:

( (r^{3n} - 1)/(r^{3}-1) ) / ( (r^{n} - 1)^{3}/(r-1)^{3} ) =

3641/11^{3} =

331/121

Letting u=r^{n} again, and putting the u's in:

( (u^{3} -1) (r-1)^{3} ) / ( (u -1)^{3}(r^{3}-1) ) = 331/121

121(u^{3} -1) (r-1)^{3} = 331 (u -1)^{3}(r^{3}-1)

121(u-1)(u^{2}+u+1) (r-1)^{3} = 331 (u -1)^{3}(r-1)(r^{2}+r+1)

121(u^{2}+u+1) (r-1)^{2} = 331 (u -1)^{2}(r^{2}+r+1)

121(u^{2}r^{2}+ur^{2}+r^{2}-2u^{2}r-2ur-2r+u^{2}+u+1)=

331(u^{2}r^{2}-2ur^{2}+r^{2}+u^{2}r-2ur+r+u^{2}-2u+1)

-210u^{2}r^{2}+783ur^{2}-210r^{2}-573u^{2}r+420ur-573r-210u^{2}+783u-210 = 0

(-210r^{2}-573r-210)u^{2} + (783r^{2}+420r+783)u + (-210r^{2}-573r-210) = 0

(-210r^{2}-573r-210)((21r+10)/(10r+21))^{2} +

(783r^{2}+420r+783)((21r+10)/(10r+21)) + (-210r^{2}-573r-210) = 0

(-210r^{2}-573r-210)(21r+10)^{2} + (783r^{2}+420r+783)(21r+10)(10r+21) +

(-210r^{2}-573r-210)(10r+21)^{2} = 0

50820r^{4} + 25410r^{3} - 152460r^{2} + 25410r + 50820 = 0

2r^{4} + r^{3} - 6r^{2} + r + 2 = 0

(r-1)^{2}(2r^{2}+5r+2)=0

(r-1)^{2}(r+2)(2r+1)=0

which has solutions r = 1 or r = -2 or r = -1/2.

Let's consider r = -2 first. We must have r^{n} = u = (21r + 10)/(10r + 21) = (-32)/(1) = -32.

So n=5.

Now, we see that:

a(r^{n}-1)/(r-1) = 11

a (-32 - 1)/(-2-1) = 11

11a = 11

a=1

which gives us the sequence as (1, -2, 4, -8, 16).

If r = -1/2, then r^{n} = u = (21r + 10)/(10r + 21) =

(-1/2)/(16) = -1/32. So n=5.

Now, we see that:

a(r^{n}-1)/(r-1) = 11

a(-1/32 - 1)/(-1/2-1) = 11

a(-33/32)/(-3/2) = 11

a(-3/32)/(-3/2)=1

a(1/16)=1

a=16

This gives us the same progression backwards --

(16, -8, 4, -2, 1)

Finally, if r=1, then r^{n} = u = (21r + 10)/(10r + 21) = 1, so there's no telling what n might be. Any "n" would make r^{n}=1.

Now, we see that:

Since r=1, the sum of the series is an=11, the sum of their squares a^{2}n=341, so a=341/11=31, but then n=11/31, which isn't an integer, so there is no solution with r=1.

In summary, the solutions are

(1, -2, 4, -8, 16)

and

(16, -8, 4, -2, 1)