Question 17.

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

A
B
C
D

Question Explanation

Text Explanation

General term = 38+(n1)17=17n+21=17(n+1)+4=17k+438 + (n-1)17 = 17n + 21 = 17(n+1) + 4 = 17k + 4

Each term is in the form of 17k+417k + 4

Least 3-digit number in the form of 17k+417k + 4 is at k=6k = 6, i.e. 106106

Highest 3-digit number in the form of 17k+417k + 4 is at k=58k = 58, i.e. 990990

106,123,140,,990106, 123, 140, \ldots, 990

990=106+17(n1)990 = 106 + 17(n-1)

n=53n = 53

Sum = 532(106+990)=53×548\frac{53}{2}(106 + 990) = 53 \times 548

Avg = 53×5485353 \times \frac{548}{53} = 548

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