**Answer : A**

**Solution :-**

**The first integer, after 60 which is divisible by 17 is 68
And the 1st integer, before 600 which is divisible by 17 is 595.
Therefore the sequence of integers between 60 and 600 which are divisible by 17 is
68, 85, 102,…,595
We have to find 68 + 85 + 102 + … + 595**

**It is an A.P with a = 68 and d = 17 and t(n) = 595
The formula to find nth term in A.P is, t(n) = a + (n – 1)d
Here, a +(n – 1)d = 595
68 + (n – 1)17 = 595
(n – 1)17 = 527
n – 1 = 31
n = 32
i.e., the 32-nd term of sequence is 595**

**We know that the sum of first n terms of the A.P series = s(n) = (n/2)[2a + (n – 1)d]
Here, s(32) = (32/2)[2(68) + (32 – 1)17]
16[136 + 31(17)]
16[136 + 527]
16[663] = 10608
Hence the required sum is 10608.**