The average of all 3-digit terms in the arithmetic progression 38, 55, 72, ..., is
Correct Answer: 548
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 = $$53\times\frac{548}{53}=548$$