June 2026 Problem of the Month

There are 4 positive integers in order from least to greatest, such that the first three terms make an arithmetic sequence, and the last three terms make a geometric sequence.

If the difference between the largest and smallest term is 30, what are the terms?

Solution to the Problem:

Here is the solution:

The answer is 18, 27, 36 and 48.

Since the four terms are distinct, and the first three terms are in arithmetic sequence, we must have the first three terms as y-b, y and y+b.   Since all the four terms are positive, it follows that y > b.

Now, the last three terms are in geometric sequence.   Thus, the common ratio is (y+b)/y, so that the 4th term is (y+b)^2/y.
Now, the difference between the last term and the first is 30, and so:

(y+b)^2/y - (y-b) = 30
or, b(3 + b/y) = 30

Since b< y, it follows that:

30 = b(3 + b/y) < 4*b, and:

30 = b(3 + b/y) > 3*b, so that:

7.5< b< 10.

The only values that satisfy the above inequality occur at b = 8, 9

Substituting b = 8 in (*), we have:

24 + 64/y = 30, or y = 32/3, which is not an integer and thus leads to a contradiction.

Substituting b = 9 in (*), we have:

24 + 81/y = 30, or y = 27, so that;
(y+b)^2/y = 36^2/27 = 48, and (y-b, y, y+b) = (18, 27, 36)

Thus, the required four terms of the given sequence are 18, 27, 36 and 48.

Click here for Dr. Kishan's excellent solution