Re: Australian Mathematics Competition
The pattern was 6, 6 + 9, 6 + 9 + 12, 6 + 9 + 12 + 15, . . .
So we have to find the 11th term, which is the sum of the first 11 terms of the arithmetic sequence with a = 6, d = 3.
T_{11} = S_{11} = \frac{11}{2}(12 + 10 \cdot 3)
= 11(6 + 15)
=11...