Question

Find the sum of all numbers between 100 and 500 which when divided by 13 leaves remainder 5.

Answer 1

Answer:

The sum of all natural numbers between 100 and 500 which are divisible by 7, is a problem of summing an AP whose first term is 105 and the last term is 497, and the common difference = 7.

Tn =497 = a +(n-1)d = 105 +(n-1)*7, or

497/7 = 71 =15+ (n-1), or n = 71–15+1 = 57

S, the sum is = (n/2)[2a+(n-1)d] = (57/2)[2*105 + (57–1)*7]

= (57/2)[210+392] = (57/2)[602]1 = 17157

Answer = 17157

Answer 2

so we should find all no.s between 100 And 500

And when divided by 13 remainder is 5

so its an AP with terms in the form of 13x+5

first term=>100/13=7.

=> 13 (7)+5=91+5=96 Which is not included

so a =13 (8)+5

=104+5

=109

a2=13 (9)+5

=117+5

=122

So d=122-109

=13

l=>500/13=38.

= 13(38)+5

= 494+5

=499

t=an=499=109+(n-1)13

n-1=499-109/13

=390/13

n=30

Sn=n/2 (a+l)

=30/2 (109+499)

=15 (608)

=9120