The sum of how many terms of the series 6 + 12 + 18 + 24 + ... is 1800 ?

This is an A.P. in which a = 6, d = 6 and S_n = 1800

Then,	n	[2a + (n - 1)d] = 1800
	2

	n	[2 x 6 + (n - 1) x 6] = 1800
	2

 3n (n + 1) = 1800

 n(n + 1) = 600

 n² + n - 600 = 0

 n² + 25n - 24n - 600 = 0

 n(n + 25) - 24(n + 25) = 0

 (n + 25)(n - 24) = 0

 n = 24

Number of terms = 24.