What is the average of first 30 multiples of 9?

Sum of the nth term = n/2 × (2a + (n – 1) × d)

n = number of terms

a = first term

d = Common difference

Sum of the 30th terms = 30/2 × (2 × 9 + (30 – 1) × 9) (n = 30, a = 9, d = 9)

= 15 × (18 + 29 × 9)

= 4185

Sum of the first 30 multiples of 9 = 4185

∴ Average of the first 30 multiples of 9 = 4185/30 = 139.5

no. of terms = ( n-1 )/2 +1

then ,{ ( 30 -1) /2 }+1 = 14.5 +1 = 15.5

then we simply multiply with 9 , then we will find answer .

So , 15.5 × 9 = 139.5