The first three terms of sequence {C(n)} are 1440, 1716 and 1848. These are obtained by multiplying the corresponding terms of two arithmetic sequences:{A(n)} and {B(n)}.
Find the 8th term of {C(n)}
1716 = 2^2 * 3 * 11 * 13
1848 = 2^3 * 3 * 7 *11
Divide by their Greatest Common Divisor, wghich is 12
This makes the terms (divided by 12)
2^3 * 3 * 15
11*13
2*7*11
Guess that the 2nd terms (ignoring the 12) are 11 and 13
Guess that the 3rd term (ignoring the 12) is 14 and 11
Then the first term (ignoring the 12) is 8 and 15, which works
So, C(n) = 12 * (5 + 3n) * (17 - 2n)
C(8) = 12 * 29 * 1 = 348
This solution might or may not be unique.
C(n) = (w + xn)(y + zn)
So we have three equations with 4 unknowns.
There are an infinite number of A(n) and B(n) which work.
But it is possible that they all lead to the same C(n)
Trial and Error (spoiler?)
