Discussion

Three numbers a, b, and c are in G.P. If abc = 27 and \(a + c\) = 10, then \(a^{2} + b^{2} + c^{2} =\)

Answer: C

\(a^{2} + b^{2} + c^{2} =\) 91.

In a G.P. \(b^{2} = ac\), so \(abc = b^{3} = 27\) gives \(b = 3\) and \(ac = 9\). Then

\[ \begin{aligned} a^{2} + c^{2} &= (a + c)^{2} - 2ac = 100 - 18 = 82 \\ a^{2} + b^{2} + c^{2} &= 82 + 9 = 91 \end{aligned} \]

(The numbers are 1, 3, 9.) 82 forgets to add \(b^{2}\), and 81, 92 and 99 come from arithmetic slips with \(ac\) or \(b^{2}\).

A general identity used here: \(a^{2} + c^{2} = (a + c)^{2} - 2ac\), with \(ac = b^{2}\) for terms in G.P.

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