Write a rule for the n th term of the geometric sequence for which a_1=11 and a_4=88
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4, 7 and 9 are mutually coprime, so you can use the Chinese remainder theorem.
Start with
Taken mod 4, the last two terms vanish and we're left with
We have , so we can multiply the first term by 3 to guarantee that we end up with 1 mod 4.
Taken mod 7, the first and last terms vanish and we're left with
which is what we want, so no adjustments needed here.
Taken mod 9, the first two terms vanish and we're left with
so we don't need to make any adjustments here, and we end up with .
By the Chinese remainder theorem, we find that any such that
is a solution to this system, i.e. for any integer
, the smallest and positive of which is 149.