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