A) AP
B) GP
C) HP
D) None of these
Correct Answer: B
Solution :
Idea We know that if x, y and 2 are in AP, then\[2y=x+z\]and for GP\[{{y}^{2}}=xz,\]for HP\[y=\frac{2x+z}{x+z}.\]Use these relations. Since, a, band c are in AP \[\therefore \]2b = a + c b,c and dare in GP \[\therefore \] c2 = bd and c, d and e are in HP \[\therefore \]\[d=\frac{2ce}{e+c}\] \[\frac{{{c}^{2}}}{b}=\frac{2ce}{c+e}\] \[\Rightarrow \]\[c(c+e)=2be\] \[\Rightarrow \]\[c(c+e)=(a+c)e\] \[{{c}^{2}}+ce=ae+ce\] \[{{c}^{2}}=ae\] Hence, a, c, e are in GP. TEST Edge Properties of series based questions are asked. To solve such types of questions students are stated that stick with definition and properties of AP, GP and HP and go through with relation between AP, GP and HP.
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