Question

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

Answer 1

Given the arithmetic progression (AP): 38, 55, 72, ...
Common difference d = 55 - 38 = 17.
The smallest 3-digit number in the AP is 106, and the largest 3-digit number is 990.
We need to find the value of:
$106 + 123 + \ldots + 973 + 990$
This is an arithmetic series with a = 106 (first term), d = 17 (common difference), and l = 990 (last term).

The formula for the sum of an arithmetic series is:
$S_n = \frac{n}{2} \cdot (a + l)$
Where n is the number of terms.
The number of terms can be calculated as:
$n = \frac{l - a}{d} + 1$

Substitute the values:
$n = \frac{990 - 106}{17} + 1 = \frac{884}{17} + 1 \approx 52.23$
The largest integer value of n that satisfies this is n = 52.

Now, use the formula for the sum:
$S_{52} = \frac{52}{2} \cdot (106 + 990) = 26 \cdot 1096 = 28496$
The average of these 52 numbers is:
$\text{Average} = \frac{S_{52}}{52} = \frac{28496}{52} = 548$

So, the average of all 3-digit terms in the arithmetic progression is indeed 548.