Question

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.

Answer 1

Sum of AP

Step-by-step explanation:

(i) Given: The first 15 multiples of 8

Find: sum  

Solution:

  n = 15

  a = 8

  d = 8

 Sum of AP = n/2 (2a + (n-1)d)

  = 15/2 (2*8 + 14(8))

  = 15/2 (16 + 112)

  = 15/2 * 128

  = 15 * 64

  = 960

(ii) The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(a) d = 3

    a = 3

    n = 40

Sum = 40/2 (6 + 39(3)) = 20 (123) = 2460

Solve b and c using the same method.

(iii) All 3 – digit natural numbers which are divisible by 13.

d = 13

 a = 104

 an = 988

 an = a + (n-1)d = 988

 988 = 104 +(n-1)13

 884 = (n-1)13

 68 = n-1

Therefore n = 69

Sum of AP = 69/2 (2(104) + 68(13))

   = 69/2 ( 208 + 884)

   = 69/2 * 1092

   = 69 * 546

   = 37674

Solve (iv) using the same method.