For any three positive real numbers a, b and c, 9(25a2+b2)+25(c2−3ac)=15b(3a+c) , then.
A.b, c and a are in G.P.
B.b, c and a are in A,P.
C.a, b and c are in A.P.
D.a, b and c are in G.P.

Question

For any three positive real numbers a, b and c, 9(25a2+b2)+25(c2−3ac)=15b(3a+c) , then. A.b, c and a are in G.P. B.b, c and a are in A,P. C.a, b and c are in A.P. D.a, b and c are in G.P.

Pawan Prajapati · Answer

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For any three positive real numbers a, b and c, 9(25a2+b2)+25(c2−3ac)=15b(3a+c) , then. A.b, c and a are in G.P. B.b, c and a are in A,P. C.a, b and c are in A.P. D.a, b and c are in G.P.

For any three positive real numbers a, b and c, 9(25a2+b2)+25(c2−3ac)=15b(3a+c) , then.
A.b, c and a are in G.P.
B.b, c and a are in A,P.
C.a, b and c are in A.P.
D.a, b and c are in G.P.

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For any three positive real numbers a, b and c, 9(25a2+b2)+25(c2−3ac)=15b(3a+c) , then. A.b, c and a are in G.P. B.b, c and a are in A,P. C.a, b and c are in A.P. D.a, b and c are in G.P.

For any three positive real numbers a, b and c, 9(25a2+b2)+25(c2−3ac)=15b(3a+c) , then. A.b, c and a are in G.P. B.b, c and a are in A,P. C.a, b and c are in A.P. D.a, b and c are in G.P.

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For any three positive real numbers a, b and c, 9(25a2+b2)+25(c2−3ac)=15b(3a+c) , then.
A.b, c and a are in G.P.
B.b, c and a are in A,P.
C.a, b and c are in A.P.
D.a, b and c are in G.P.