Question

If the angles $[A,B,C]$ of a triangle are in A.P. and the sides $[a,b,c]$ opposite to these angles are in G. P. then$ [{a^2},{b^2},{c^2}]$ are in
A. A.P.
B. H.P.
C. G.P.
D. None of these

Answer 1

Hint: The hint for the problem is to use the law of sines. Given that $[A,B,C]$ are in A.P., \[\;a2{\rm{ }} = {\rm{ }}b2{\rm{ }} + {\rm{ }}c2\] and so G. P. also contains these values$[\left( {A - B = G,{\rm{ }}B - C = H} \right)].$
According to the sine rule, if a, b, and c are a triangle's side lengths and a, b, and c are its angles, with an opposite a, etc., then$[asinA = bsinB = csinC].$
When two angles and one side of a triangle are known, a process known as triangulation can be used to calculate the other sides using the law of sines. When two sides and one of the non-enclosed angles are known, it can also be used.
When we are provided either a) two angles and one side or b) two sides and an excluded angle, the sine rule is used. When we have either a) three sides or b) two sides and the included angle, we apply the cosine rule.

Complete step by step solution: As the angle $[A,B,C]$are in A.P.,
So, $[2B = A + C]$
$ = > 3B = A + B + C$
$ = > 3B = 180^\circ $
Then, the value of B becomes,
$ = > B = 60^\circ $
$ = > \cos B = \dfrac{1}{2}$
$ = > \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}} = \dfrac{1}{2}$
$ = > {c^2} + {a^2} - {b^2} = ac$
So, the a,b,c are in G.P
$ = > {c^2} + {a^2} - {b^2} = {b^2}$
$ = > {c^2} + {a^2} = 2{b^2}$
Hence, the$ {a^2},{b^2},{c^2}$are in A.P.

So, Option ‘A’ is correct

Note: The notes for the problem are as follows:
If the angles$ [A,B,C] $of a triangle are in A.P. and the sides$ [a,b,c] $opposite to these angles are in G. P. then$ [{a^2},{b^2},{c^2}]$ are in A.
The notes for the problem are as follows:
If the angles $[A,B,C]$ of a triangle are in A.P. and the sides$ [a,b,c]$ opposite to these angles are in G. P. then$ [a{x^2} + b{y^2} + cz = g\left( {a - x} \right)]$, where $[g] $is an inverse function of$ [P]$ such that $[z = - g\left( a \right)].$
In this case, $[z = - 9/12]$ because $[9/12 \times \left( { - 1} \right) = - 9]$ which leaves on each side (since $[3]$ is not divisible by $[6]).$