If In(a+c),In(c-a),In(a-2b+c)are in A.P., then
a,b,c are in A.P
a2,b2,c2 are in A.P
a,b,c are in G.P
a,b,c are in H.P
Explanation for correct option:
Find the relation:
In(a+c),In(c–a),In(a–2b+c) are in A.P.
So,
2In(c–a)=In(a+c)+In(a–2b+c)[∵2b=a+c]⇒(c–a)2=(a+c)(a–2b+c)[∵nloga=logan,loga+logb=logab]⇒c2–2ac+a2=a2+ac–2ab–2bc+ac+c2⇒-2ac=-2ab+2ac–2bc⇒2ac=ab+bc⇒2ac=b(a+c)⇒2b=(a+c)ac
So a,b,c are in HP.
Hence, the correct option is D.
If sin(B+C−A),sin(C+A−B),sin(A+B−C) are in A.P., then cot A, cotB, cotC are in
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