Archimedes Laboratory Project™
Puzzles & Mental Activities that Enhance Critical and Creative Thinking Skills

	Prove that if 1/(a+b), 1/(a+c) and 1/(b+c) are the consecutive terms of an arithmetic sequence, then a2, b2 and c2 are also in arithmetic progression (and vice versa). Démontrez que si 1/(a+b), 1/(a+c) et 1/(b+c) sont les termes d’un séquence arithmétique, alors a2, b2 et c2 sont eux aussi en progression arithmétique (et réciproquement). Dimostrare che se 1/(a+b), 1/(a+c) e 1/(b+c) sono i termini consecutivi di una successione aritmetica, allora a2, b2 e c2 sono anch’essi in progressione aritmetica (e viceversa). You can comment and discuss this puzzle on our FaceBook or Blogger web page. › Click to show/hide the solution Solution: If N1, N2, and N3 are consecutive terms of an arithmetic sequence, then: 2·N2 – (N1 + N3) = 0 Likewise: 2/(a+c) – [1/(a+b)+1/(b+c)] = 0 Simplifying: 1. 2b2 – (a2 + c2)/(a+b)(b+c)(a+c) = 0 2. 2b2– (a2 + c2) = 0 Then 2b2 = a2 + c2 Subconsequently a2, b2 and c2 are also in arithmetic progression. Please spread the word about the Sunday Puzzle!

Text and images by Gianni A. Sarcone 
Labels: SUNDAY PUZZLE, puzzles to solve, recreational mathematics, progression
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Sunday Puzzle by Gianni A. Sarcone is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.
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