A) A.P.
B) GP.
C) H.P.
D) None of these
Correct Answer: B
Solution :
Here the given condition \[({{a}^{2}}+{{b}^{2}}+{{c}^{2}}){{p}^{2}}-2p(ab-bc-cd)+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}\le 0\]\[\Rightarrow \,\,\,{{(ap-b)}^{2}}+{{(bp-c)}^{2}}+{{(cp-d)}^{2}}\le 0\] Since the squares cannot be negative \[\Rightarrow \,\,\,{{(ap-\,b)}^{2}}+{{(bp\,-c)}^{2}}+{{(cp-d)}^{2}}\le \,\,0\] \[\Rightarrow \,\,\,\frac{1}{p}=\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\] \[\therefore \] a, b, c, d are in GP.You need to login to perform this action.
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