Question

Find number of terms in the arithmetic progression 6, 15, 24, 33, ... such that sum
of the series is 2850?

Answer 1

Answer:

$a = 6 \\ d = 9 \\ sn = 2850 \\ sn = \frac{n}{2} (2a + (n - 1)d) \\ 2850 = \frac{n}{2} (12 + 9n - 9) \\ 2850 = \frac{n}{2} (3 + 9n) \\ 5700 = 3n + 9n {}^{2} \\ 9n {}^{2} + 3n - 5700$