Question

37. The sum of all natural numbers from 100
to 300, which are divisible by 5, is
a) 10200
b) 30000 (#)(€)
c) 8200
d) 2200
​

Answer 1

Answer:

c) 8200

Step-by-step explanation:

Sum of numbers divisible by 5 = 100 + 105 + 110 + 115 + ... + 300

Total numbers = 41

Sn = n/2[2a + (n – 1)d] ( sum of arithmetic progression)

= 41/2[200 + 40 × 5]

= 41/2[400]

= 41 × 200

= 8200

HOPE THIS HELPS YOU......

Answer 2

Answer:

Given,

A = 100

D = 5

L = 300

=> An = a + (n-1)d

=> 300 = 100 + (n-1)5

=> 200 = 5n - 5

=> 205 = 5n

=> n = 41 terms

Sn = n/2 [2a + (n-1)d]

= 41/2 [2 x 100 + 40 x 5]

= 41/2 [ 200 + 200]

= 41/2 x 400

= 41 x 200

= 8200

Sum = 8,200