November 2024 Problem of the Month

Sequence Problem
Submitted by K. Sengupta of Calcutta, INDIA

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Solution to the Problem:

The four terms are: 18, 27, 36 and 48.

Since the four terms are distinct, and the first three terms are in arithmetic sequence, let the first three terms be 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 b(3 + b/y) = 30, 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 b(3 + b/y) = 30, 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.

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Here is Kamal Lohia's solution:

Let the four positive integers be: a-d, a, a+d, a-d+30
such that (a+d)² = a(a-d+30)
=> d² + 3ad - 30a = 0
=> d = ½{-3a ±√(9a²+120a)}

Now, 9a² + 120a should be a perfect square.
That's (3a)² + 2(3a)(20) + (20)² - 400 should be a perfect square.
That's (3a+20)² - 400 = k²
=> (3a+20-k)(3a+20+k) = 2•200 or 4•100 or 8•50 or 10•40 or 20•20 as both terms must be even.

Simplifying, we get a can be 27, 3 or 0 leaving out 2nd and 4th possibility as they don't lead in an integer value for a.

Now, checking the corresponding values of d respectively, we get d will be 9, 6 or 0 respectively.

Discarding the last two cases, for not getting the 4 positive integers, we get the four numbers as; 18, 27, 36, 48.

Correctly solved by: