The sum of the first n terms of the arithmetic progression 5, 11, 17, 23, 29, 35, ... is
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Correct Answer: Option B
Explanation:
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)
= 6n[sup]2[/sup]/2 + 4n/2
= 32 + 2n
= n(3n + 2)
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)
= 6n[sup]2[/sup]/2 + 4n/2
= 32 + 2n
= n(3n + 2)