Question

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

Answer 1

First term, a = 6
Common difference, d = 15-6 = 9
Let the number of terms be n
$sum = 2850 \\ \frac{n}{2} \times (2a + (n - 1) \times d) = 2850 \\ \frac{n}{2} \times (2 \times 6+ (n - 1) \times 9) = 2850 \\ n \times (9n + 3) = 5700 \\ 9 {n}^{2} + 3n - 5700 = 0 \\taking \: \: only \: \: + ve \: value \: \: for \: \: n \\ n = \frac{ - 3 + \sqrt{ {( - 3)}^{2} - 4 \times 9 \times - 5700 } }{2 \times 9} \\ = \frac{ - 3 + 453}{18} = 25$

Answer 2

Answer:

First term, a = 6

Common difference, d = 15-6 = 9

Let the number of terms be n