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 Sum from Arithmetic and Geometric (Posted on 2016-03-24)
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

 No Solution Yet Submitted by K Sengupta No Rating

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 Possible Solution | Comment 1 of 5
Call the difference d so 20A=6B-d, C=6B+d
Call the ratio r so that 20A=6B/r, C+1=6Br
Equating the 20A terms gives
6B=dr/(r-1)
Equating the C terms gives
6B=(d+1)/(r-1)
Equating these 6B terms gives
r=(d+1)/d
so r is determined by d but so are A,B,C as
A=d²/20
B=d(d+1)/6
C=d(d+2)
C+1=(d+1)²
A will only be an integer if d is a multiple of 10.
B will only be an integer if d is a multiple of 3 or 1 less than a multiple of 3.
So d must be 20, 30, 50, 60, 80, 90, etc...
and since A+B+C=(73d²+130d)/60 it's clear the smallest 6 valid values of d will give the 6 smallest values of A+B+C
` d  20A   6B    C  C+1 A+B+C20  400  420  440  441   53030  900  930  960  961  116050 2500 2550 2600 2601  315060 3600 3660 3720 3721  451080 6400 6480 6560 6561  796090 8100 8190 8280 8281 10050-10  100   90   80   81   100-30  900  870  840  841  1030-40 1600 1560 1520 1521  1860-60 3600 3540 3480 3481  4250`
Edit:  I made the error of assuming d had to be positive.  There are other small solutions with negative d as appended to the table.  Bolding changed to the actual 6 smallest.  I also fixed a small typo.

Edited on March 24, 2016, 8:17 pm
 Posted by Jer on 2016-03-24 13:22:49

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