Given,
An AP is required of all integers between 84 and 719, which are multiples of 5
To find : the sum of all integers between 84 and 719, which are multiples of 5
So,
The sequence is 85, 90, 95….715
It is an AP whose first term is 85 and d is 5
Hence,
The sum is given by the formula s = \(\frac{n}{2}\)(2a+(n-1)d)
Now,
For the finding number of terms, the formula is
n = 127
Substituting n is the sum formula we get,
s = \(\frac{127}{2}\)(2 x 85 + (126) x 5)
s = 50800.
