Question 13

# The average of all 3-digit terms in the arithmetic progression 38, 55, 72, ..., is

Solution

General term = 38 + (n-1)17 = 17n + 21 = 17(n+1) + 4 = 17k + 4

Each term is in the form of 17k + 4

Least 3-digit number in the form of 17k + 4 is at k = 6, i.e. 106

Highest 3-digit number in the form of 17k + 4 is at k = 58, i.e. 990

106, 123, 140,..........., 990

990 = 106 + 17(n-1)

n = 53

Sum = $$\frac{53}{2}\left(106+990\right)=53\times548$$

Average = $$53\times\frac{548}{53}=548$$