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306,315,324,333.......693
First Multiple After 300 = 306.
Last Multiple Before 700 = 693.
a = 306, d = 9, an = 693.
We know that sum of n terms of an AP an = a + (n - 1) * d
693 = 306 + (n - 1) * 9
693 = 306 + 9n - 9
693 = 297 + 9n
693 - 297 = 9n
396/9 = n
n = 44.
Now,
s44 = n/2(a + l)
= 44/2(306 + 693)
= 22(999)
= 21978.
Therefore the required sum is 21978.
Hope this helps!