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Find the sum of all the natural numbers from 1 and 400, inclusive, that are not divisible by 11.

Sagot :

Jmmoralesidn
Firstly try to find the number between 100 and 400 which are actually divisible by 13 so

104,117,130………. 390

If you see the above arrangement the difference between numbers are common so we can apply arithmetic progression.

By arithmetic progression, last term can be denoted by

l= a+(n-1)d

Where l=390

a=104

d=13

n=?

390=104+(n-1)13

13n=286+13

n= 299/13

n= 23

So total number of terms =23

By using arithmetic progression sum of all A.P terms = (n/2)(a+l)

Sum = (23/2)(104+390)

=(23/2)*(494)

= 5,681

So answers is 5,681.
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