Question

The sum of all natural numbers between 1 to 70 divisible by 7??​

Answer 1

Answer:

We observe that 14 is the first two-digit number divisible by 7 and 98 is the last two-digit number divisible by 7.

Thus, we have to determine the number of terms in the sequence.

14,21,28,...,98

Clearly, it is an A.P. with first term =14 and common difference =7 i.e. a=14 and d=7.

Let this be the n

th

term in this A.P.

Then, n

th

term =98

⟹14+(n−1)×=98

⟹14+7n−7=98

⟹7n=91⟹n=13

Hence, there are 13 numbers of two digits which are divisible by 7.

Answer 2

Sum of first n natural numbers = n(n+1)/2

Here, n = 70

Sum = 70(70+1)/2

= 2485

So, 2485 is divisible by 7

On dividing we get 355