Sum of terms in AP = \(\frac{n}{2}[2(a) + (n - 1)d]\)
Number of terms in multiples = \(\frac{(L - F)}{multiple} + 1\)

Number of multiples of 3 = \(\frac{(99 - 3)}{3} + 1 = 33\)
Number of multiples of 6 = \(\frac{(198 - 6)}{6} + 1 = 33\)

Sum of multiples of 3 = \(\frac{33}{2}[2(3) + (33 - 1)3] = \frac{33}{2}[6 + 96] = 1,683\)
Sum of multiples of 6 = \(\frac{33}{2}[2(6) + (33 - 1)6] = \frac{33}{2}[12 + 192] = 3,366\)

Clearly, Sum of multiples of 6 is twice the sum of multiples of 3
Therefore, s = 2t