JAMB - Mathematics (2011 - No. 18)
Find the sum of the first 18 terms of the series 3, 6, 9,..., 36.
505
513
433
635
Erläuterung
3, 6, 9,..., 36.
a = 3, d = 3, i = 36, n = 18
Sn = \(\frac{n}{2}\) [2a + (n - 1)d
S18 = \(\frac{18}{2}\) [2 x 3 + (18 - 1)3]
= 9[6 + (17 x 3)]
= 9 [6 + 51] = 9(57)
= 513
a = 3, d = 3, i = 36, n = 18
Sn = \(\frac{n}{2}\) [2a + (n - 1)d
S18 = \(\frac{18}{2}\) [2 x 3 + (18 - 1)3]
= 9[6 + (17 x 3)]
= 9 [6 + 51] = 9(57)
= 513