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

a(1 + r + r² + ... + rⁿ-1) = 11
a²(1 + r² + r⁴ + ... + r^2n-2) = 341
a³(1 + r³ + r⁶ + ... + r^3n-3) = 3641

Using the standard geometric progression formulae,

a (rⁿ - 1)/(r-1) = 11
a² (r²ⁿ - 1)/(r²-1) = 341
a³ (r³ⁿ - 1)/(r³-1) = 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:

(r²ⁿ - 1) (r - 1)² / ( (r² - 1)(rⁿ - 1)²) =
341/112 =
31/11

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

We have

(u² - 1) (r - 1)² / ( (r² - 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

( (r³ⁿ - 1)/(r³-1) ) / ( (rⁿ - 1)³/(r-1)³ ) =
3641/11³ =
331/121

Letting u=rⁿ again, and putting the u's in

( (u³ -1) (r-1)³ ) / ( (u -1)³(r³-1) ) = 331/121
121(u³ -1) (r-1)³ = 331 (u -1)³(r³-1)
121(u-1)(u²+u+1) (r-1)³ = 331 (u -1)³(r-1)(r²+r+1)
121(u²+u+1) (r-1)² = 331 (u -1)²(r²+r+1)
121(u²r²+ur²+r²-2u²r-2ur-2r+u²+u+1)= 331(u²r²-2ur²+r²+u²r-2ur+r+u²-2u+1)
-210u²r²+783ur²-210r²-573u²r+420ur-573r-210u²+783u-210 = 0
(-210r²-573r-210)u² + (783r²+420r+783)u + (-210r²-573r-210) = 0
(-210r²-573r-210)((21r+10)/(10r+21))² + (783r²+420r+783)((21r+10)/(10r+21)) + (-210r²-573r-210) = 0
(-210r²-573r-210)(21r+10)² + (783r²+420r+783)(21r+10)(10r+21) + (-210r²-573r-210)(10r+21)² = 0
50820r⁴ + 25410r³ - 152460r² + 25410r + 50820 = 0
2r⁴ + r³ - 6r² + r + 2 = 0
(r-1)²(2r²+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 rⁿ = u = (21r + 10)/(10r + 21) = (-32)/(1) = -32. So n=5.

Now, we see that

a(rⁿ-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 rⁿ = u = (21r + 10)/(10r + 21) = (-1/2)/(16) = -1/32. So n=5.

Now, we see that

a(rⁿ-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 rⁿ = u = (21r + 10)/(10r + 21) = 1, so there's no telling what n might be. Any "n" would make rⁿ=1.

Now, we see that

Since r=1, the sum of the series is an=11,
the sum of their squares a^2n=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)