Discussion
If \(a, b, c\) are in A.P., \(b, c, d\) are in G.P. and \(1/c, 1/d, 1/e\) are in A.P., then \(a, c, e\) are in:
Answer: B
Given \(2b = a+c\), \(c^2=bd\), and \(c,d,e\) are in H.P. so \(d = \frac{2ce}{c+e}\). From \(c^2=bd\), we have \(d=c^2/b\). So \(c^2/b = \frac{2ce}{c+e} \Rightarrow c/b = \frac{2e}{c+e} \Rightarrow c(c+e) = 2be\). From the first condition, \(b=(a+c)/2\). Substitute this: \(c(c+e)=2(\frac{a+c}{2})e \Rightarrow c^2+ce = (a+c)e = ae+ce \Rightarrow c^2 = ae\). This is the condition for a, c, e to be in G.P.
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