Each of A, B and C is a positive integer such that:
20*A, 6*B and C are in arithmetic sequence, and:
20*A, 6*B and C+1 are in geometric sequence

Find the six smallest values of A+B+C

From the arithmetic sequence: a, a+c, a+2c.

From the geometric sequence: a, ad, ad^2-1.

a+c=ad, so d=(a+c)/a. a+2c=ad^2-1, so a((a+c)/a)^2-(a+2c)=1, simplifying neatly to a=c^2, when d=1/c+1.

So the series are: c^2,c^2+c,c^2+2c, and c^2, c^2(1/c+1) =c^2+c, and c^2(1/c+1)^2 = (c+1)^2. These solve for all c, so Excel produces a chart of values, and it is just necessary to check for those compliant with the problem.

Looking into it a bit further though, we can derive two separate solution families from the data given:

A = 5(9n^2+6n+1),   B = 5(30n^2+19n+3),   c = -10(3n+1), and

A = 45n^2,   B = 5(30n^2-n),   c = -30n, 

(with a small 'c' in each case)

The first family gives for n= 0,-1,1, and -2, c= -10,20,-40,and 50, solutions 1,2,5,and 6:

n,c,A,B,C,(A+B+C), solution number

-2 50 125 425 2600 3150 6
-1 20 20 70 440 530 2
0 -10 5 15 80 100 1
1 -40 80 260 1520 1860 5

The second family gives for n=1, -1 c=-30, and 30, solutions 3 and 4.

n,c,A,B,C,(A+B+C), solution number

-1 30 45 155 960 1160 4
1 -30 45 145 840 1030 3

On checking, these values agree with those given by Charlie's program.

Edited on March 24, 2016, 11:17 pm

Posted by broll on 2016-03-24 23:07:43