Question

If a, b, c are in G.P and loga-log2b, log2b-log3c and log3c-loga are in A.P, then a, b, c are the lengths of the sides of a triangle which is
$
  (a){\text{ Acute angled}} \\
  (b){\text{ Obtuse angled}} \\
  (c){\text{ Right angled}} \\
  (d){\text{ Equilateral}} \\
$

Answer 1

Hint: In this question three numbers a, b and c are said to be in G.P. Numbers of a series are said to be in G.P if the common ratio that is the ratio of consecutive terms remains constant throughout. Three different numbers are said to be A.P, so numbers of a series are said to be A.P if the common difference, that is the difference between the consecutive numbers remains constant throughout. Use these properties of A.P and G.P along with the cosine property of triangle to get the answer.

Complete step-by-step answer:
It is given that a, b, c are in G.P
Therefore according to property of G.P common ratio (r) should be equal.
$ \Rightarrow r = \dfrac{b}{a} = \dfrac{c}{b}$…………………….. (1)
Now it is also given that $\left( {\log a - \log 2b} \right),\left( {\log 2b - \log 3c} \right),\left( {\log 3c - \log a} \right)$ are in A.P.
So according to the property of A.P common difference (d) should be equal.
$ \Rightarrow d = \left[ {\left( {\log 2b - \log 3c} \right) - \left( {\log a - \log 2b} \right)} \right] = \left[ {\left( {\log 3c - \log a} \right) - \left( {\log 2b - \log 3c} \right)} \right]$
Now simplify the above equation we have,
$ \Rightarrow \left[ {2\log 2b - \log a - \log 3c} \right] = \left[ {2\log 3c - \log a - \log 2b} \right]$
$ \Rightarrow 3\log 2b = 3\log 3c$
Now divide by 3 throughout and on comparing we have,
$ \Rightarrow 2b = 3c$
$ \Rightarrow b = \dfrac{3}{2}c$
Now from equation (1) substitute the value of b we have,
 $ \Rightarrow \dfrac{{\dfrac{3}{2}c}}{a} = \dfrac{c}{{\dfrac{3}{2}c}}$
Now simplify it we have,
$ \Rightarrow a = \dfrac{9}{4}c$
Now it is given a, b and c are the lengths of the triangle.
And we see from above calculation that (a) is the greatest side.
Therefore according to cosine property in a triangle we have,
$\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$ (Where A is one of the angle of the triangle).
Now substitute the value of a, b and c in above equation we have,
$ \Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \dfrac{{{{\left( {\dfrac{3}{2}c} \right)}^2} + {c^2} - {{\left( {\dfrac{9}{4}c} \right)}^2}}}{{2\left( {\dfrac{3}{2}c} \right)c}} = \dfrac{{\dfrac{9}{4} + 1 - \dfrac{{81}}{{16}}}}{3}$
Now simplify the above equation we have,
$ \Rightarrow \cos A = \dfrac{{\dfrac{9}{4} + 1 - \dfrac{{81}}{{16}}}}{3} = \dfrac{{36 + 16 - 81}}{{3 \times 16}} = \dfrac{{ - 29}}{{48}} < 0$
So as we see that cosine of A is less than zero therefore angle A should be greater than 90 degree.
Therefore A must be Obtuse angle.
Therefore triangle A is an obtuse triangle.
Hence option (B) is correct.

Note: Whenever we face such types of problems the key concept is simply to have the understanding of obtuse and acute angles triangle. If cosine property in a triangle is giving the angle less than zero then it means that the angle must be greater than 90 degrees and opposite of this if cosine property is giving a positive sign. If angle is greater than 90 degrees then it must be an obtuse angle. Use these properties along with basic A.P and G.P identities to get answers for such problems.