Question

The first a
10. Find the sum of all natural numbers between 100 and 500 which are divisible by 7​

Answer 1

Answer:

The numbers lying between 100 and 500 which are divisible

 by 8 are

104,112,120,128,136,...,496

a2−a1=112−104=8

a3−a2=120−112=8

∵a3−a2=a2−a1=8

Therefore, the series is in AP

Here, a=120,d=8 and an=496

We know that,

an=a+(n−1)d

⇒496=104+(n−1)8

⇒496−104=(n−1)8

⇒392=(n−1)8

⇒49=(n−1)

⇒n=50

Now, we have to find the sum of this AP

Sn=2n[2a+(n−1)d]

⇒S50=2