CAT 2022 Slot 3 · Quant · Question 17

The progression starts at 38 and rises by 55-38=17 each time.

A general term is 38+17k for k=0,1,2,\ldots

The three-digit terms are those between 100 and 999 inclusive.

For the lower end, 38+17k\ge 100 gives 17k\ge 62, so k\ge 3.65 and the first whole value is k=4.

That gives the term 38+68=106.

For the upper end, 38+17k\le 999 gives 17k\le 961, so k\le 56.5 and the last whole value is k=56.

That gives the term 38+952=990.

The three-digit terms therefore run from 106 to 990, still in steps of 17 — so they form an arithmetic progression of their own.

The average of any arithmetic progression is the average of its first and last terms.

That is \frac{106+990}{2}=\frac{1096}{2}=548.

Shortcut. There is no need to count how many terms there are: the mean of an evenly spaced list is just the midpoint of its ends. Finding 106 and 990 is the whole task.