The sum of all natural numbers between 1 to 70 divisible by 7??
Answers
Answered by
2
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.
Answered by
1
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
Similar questions