A) a, b and c are in G.P.
B) b, c and a are in G.P.
C) b, c and a are in A.P.
D) a, b and c are in A.P.
Correct Answer: C
Solution :
\[{{(15a)}^{2}}+{{(3b)}^{2}}+{{(5c)}^{2}}-(15a)(5c)-(15a)(3b)\] \[-(3b)(5c)=0\] \[\frac{1}{2}[{{(15a-3b)}^{2}}+{{(3b-5c)}^{2}}+{{(5c-15a)}^{2}}]=0\] it is possible when 15a = 3b = 5c \[\therefore \] \[b=\frac{5c}{3},\,a=\frac{c}{3}\] \[a+b=2c\] \[\Rightarrow \] \[b,c,\,a\]in \[A.P.\]You need to login to perform this action.
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