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)}
Define the terms of A(n) to be a + (n-1)*x for integer n, and similarly define the terms of B(n) to be b + (n-1)*y for integer n. Then the terms of C(n) are ab + (n-1)*(ay+bx) + (n-1)^2*xy.
The first three terms of C(n) are given, so:
C(1) = ab = 1440
C(2) = ab + 1*(ay+bx) + 1*xy = 1716
C(3) = ab + 2*(ay+bx) + 4*xy = 1848
Treat this as a system of linear equations with variables ab, ay+bx, and xy. Then solve to find:
ab = 1440
ay+bx = 348
xy = -72
Then C(8) = ab + 7*(ay+bx) + 49*xy = 1440 + 7*348 + 49*(-72) = 348.
Solution
