If a, b, c are in A.P., and a², b²,c² are in H.P. (a ≠ b ≠ c) then:
(A) a 2b,c are in G.P
(B) a, b, care in H.P
(C) a, b, -c/2 are in G.P.
(D) -a/2 , b, c are in GP
(More than one are correct, and please try to explain it)
Answers
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3
Answer:
Since a, b, c are in AP
2b = a+c
(2bx +1)2 = (22bx +2)
= 2(a+c)x +2
= 2ax+1 × 2cx+1
=> 2ax +1 , 2bx +1, 2cx +1 are in GP.
Hence option (4) is the answer.
hope it will help you
Step-by-step explanation:
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