JAMB - Mathematics (2009 - No. 16)
The sum of the first n terms of the arithmetic progression 5, 11, 17, 23, 29, 35, ... is?
n(3n - 0.5)
n(3n + 2)
n(3n + 2.5)
n(3n + 5)
Explication
a = 5, d = 6, n = n
Sn = n/2(2a + (n-1)d)
= n/2(2(5) + (n-1)6)
= n/2(10 + 6n-6)
= n/2(6n+4)
= 6n2/2 + 4n/2
= 32 + 2n
= n(3n + 2)
Sn = n/2(2a + (n-1)d)
= n/2(2(5) + (n-1)6)
= n/2(10 + 6n-6)
= n/2(6n+4)
= 6n2/2 + 4n/2
= 32 + 2n
= n(3n + 2)