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in Logarithm by (24.0k points)
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If x, y, z are in G.P. and (log x – log 2y), (log 2y – log 3z) and (log 3z – log x) are in A.P., then x, y, z are the lengths of the sides of a triangle which is : 

(a) acute angled 

(b) equilateral 

(c) right angled 

(d) obtuse angled

1 Answer

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Best answer

(d) obtuse angled

x, y, z are in G.P. ⇒ y2 = xz               ...(i) 

(log x – log 2y), (log 2y – log 3z) and (log 3z – log x) are in A.P. ⇒ 2(log 2y – log 3z) = (log x – log 2y) + (log 3z – log x) 

⇒ 3 log 2y = 3 log 3z ⇒ log 2y = log 3z ⇒ y = \(\frac{3}{2}\)

∴ Putting the value of y in (i), we have

\(\big(\frac{3}{2}z\big)^2\) = xz ⇒ x = \(\frac{9}{4}\)z.

Now, by the cosine rule of triangles,

Cos A = \(\frac{y^2+z^2-x^2}{2yz}\)

where x is the length of the side opposite ∠A.

 cos A is less than 0, i.e, negative, ∠A is obtused and the triangle is obtuse angled.

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