Find the sum of
(i) The first 15 multiples of 8
(ii) The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.
(iii) All 3 – digit natural numbers which are divisible by 13.
(iv) All 3 – digit natural numbers, which are multiples of 11.
(i) The first 15 multiples of 8
a = 8, d = 8
S15= [2(8) = 14(8)]
= 120 [1 + 7]
= 960
(ii) The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.
(a) a = 3, d = 3
S40= [2(a) = 39(d)]
= 20 [2(3) = 39(3)]
= 60 (41) = 2460
(b) a = 6, d = 5
S40= [2(5) + 39(5)]
= 100 [41]
= 4100
(iii) All 3 – digit natural numbers which are divisible by 13.
a = 6, d = 6
S40= [2(6) + 39(6)]
= 120 (41) = 4920
(iv) All 3 – digit natural numbers, which are multiples of 11.
a = 110, d = 11, l = 990
n = + 1
= + 1
= =
= 81
S81= [2(110) + 80(11)]
= 8910 [1 + 4]
= 44550