November 2024
Problem of the Month

Sequence Problem
Submitted by K. Sengupta of Calcutta, INDIA



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

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



Solution:

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.



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:

1. Kamal Lohia Holy Angel School,
Hisar, Haryana, India
2. Yuvraj Sharma Holy Angel School (alumnus) and Shaheed sukhdev college of business studies
Hisar, Haryana, India
3. Dr. Hari Kishan D.N. College,
Meerut, Uttar Pradesh, India
4. Sudhir Bavdekar Mumbai, India
5. Colin (Yowie) Bowey Beechworth, Victoria, Australia
6. Davit Banana Istanbul, Turkey
7. Kelly Stubblefield Mobile, Alabama, USA


Send any comments or questions to: David Pleacher