Answer:

The series is 85, 90, 95, …, 715

‘n’ terms in the AP

a = 85, d = 90-85 = 5, a_n = 715

a_n = a + (n-1)d

715 = 85 + (n-1)5

715 = 85 + 5n – 5

5n = 715 – 85 + 5

5n = 635

n = 635/5

n = 127

By using the formula,

Sum of n terms,

S = n/2 [a + l]

S = 127/2 [85 + 715]

S = 127/2 [800]

S = 127 [400]

S = 50800

∴ The sum of all integers between 84 and 719, which are multiples of 5 is 50800.