Question -
Answer -
The series is 85, 90,95, …, 715
Let there be ‘n’ termsin the AP
So, a = 85, d = 90-85= 5, an = 715
an = a+ (n-1)d
715 = 85 + (n-1)5
715 = 85 + 5n – 5
5n = 715 – 85 + 5
5n = 635
n = 635/5
= 127
By using the formula,
Sum of n terms, S =n/2 [a + l]
= 127/2 [85 + 715]
= 127/2 [800]
= 127 [400]
= 50800
∴ The sum of all integersbetween 84 and 719, which are multiples of 5 is 50800.