ICSE & CBSE Board Question Based on Arithmetic Progression of RS Aggarwal

Here you have to find the

(i) sum of all natural numbers

(ii) Between two numbers which is given

This is the Question Number 10 Of Exercise 11 C of RS Aggarwal Solution.

Deepak BoraNewbie

# Find the sum of all natural numbers between 100 and 500 which are divisible by 7.

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105,+112……+497

let a= 105

d=7

an= 497

to find n

a+(n-1)d= an

105+(n-1)7= 497

(n-1)7= 497-105

n-1= 392/7

n-1= 56

n=57

now sum we know..

Sn=n/2 ( an+a)

=57/2 (497+105)

57/2(602)

57 x 301

=17,157