Answer : (c) 5220
The numbers between 107 and 253 divisible by 5 are
110, 115, 120, .......... , 245, 250.
This is an A.P with first term (a) = 110 and common difference (d) = 5.
Let the last term be the nth term.
∴ Tn = a + (n – 1)d
⇒ 110 + (n – 1) × 5 = 250
⇒ 5n = 250 – 105 = 145
⇒ n = 29.
∴ Required sum = \(\frac{n}{2}\) (a + Tn) = \(\frac{29}{2}\)(110 + 250)
= \(\frac{29}{2}\) × 360 = 5220