First 10 multiples of 6 are:

6, 12, 18, 24, . . . . . 60

To calculate sum using arithmetic progression, we need first term (a), common difference (d) and number of terms:

here n = 10, a = 6

and d = 12 - 6 = 6

Now, S = (n/2)[2a + (n - 1)d]

S = (10/2)[(2x6) + (10 - 1)6]

S = 5[12 + 54]

S = 5x66

S = 330

Sum of first 10 multiples of 6 is 330