A) A.P.
B) G.P.
C) H.P.
D) none of these
Correct Answer: B
Solution :
\[({{a}^{2}}+{{b}^{2}}+{{c}^{2}}){{p}^{2}}-2(ab+bc+ca)p\] \[+({{b}^{2}}+{{c}^{2}}+{{d}^{2}})\le 0\]...(i) LHS\[=({{a}^{2}}{{p}^{2}}-2abp+{{b}^{2}})\] \[+({{b}^{2}}{{p}^{2}}+2bcp+{{c}^{2}})+({{c}^{2}}{{p}^{2}}-2cdp+{{d}^{2}})\] \[={{(ap-b)}^{2}}+{{(bp-c)}^{2}}+{{(cp-d)}^{2}}\ge 0\]...(ii) since the sum of squares of real numbers is non-negative \[\therefore \]from equation (i) and (ii), we get \[\Rightarrow \] \[{{(ap-b)}^{2}}+{{(bp-c)}^{2}}+{{(cp-d)}^{2}}=0\] \[\Rightarrow \] \[ap-b=0bp-c=cp-d\] \[\Rightarrow \] \[\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p\] \[\therefore \] \[a,b,c,d\]are in G.P.
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