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+1 vote
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in Binomial theorem by (46.6k points)

If a, b, c are in GP and loga − log2b, log2b − log3c and log3c − loga are AP, then a, b, c are the lengths of the sides of a triangle which is

(A)  acute angled

(B)  obtuse angled

(C)  right angled

(D)  no triangle will be formed

1 Answer

+1 vote
by (52.7k points)
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Best answer

Correct option  (B) obtuse angled

We have 2(log2b − log3c) = loga − log2b + log3c − loga ⇒ 3(log2b − log3c) = 0 ⇒ 2b = 3c

Further, b2 = ac ⇒ 9c2 /4 = ac ⇒ c = 4a/9

Thus, a = 9c/4 and b = 3c/2 ⇒ a : b : c = 9/4 : 3/2 : 1 = 9 : 6 : 4

Clearly, sum of any two are greater than third, so they form a triangle.

Also, cos A = 42 + 62 - 92/2.4.6 = 29/48 ⇒ is obtuse angle.

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