If the angles $$A < B < C$$ of a triangle are in A. P., then
A.
$${c^2} = {a^2} + {b^2} - ab$$
B.
$${b^2} = {a^2} + {c^2} - ac$$
C.
$${c^2} = {a^2} + {b^2} $$
D.
None of these
Answer :
$${b^2} = {a^2} + {c^2} - ac$$
Solution :
$$\eqalign{
& A + C = 2B{\text{ and }}A + B + C = {180^ \circ }{\text{ so, }}B = {60^ \circ } \cr
& \therefore \cos {60^ \circ } = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} \cr
& \Rightarrow {b^2} = {a^2} + {c^2} - ac \cr} $$
Releted MCQ Question on Algebra >> Sequences and Series
Releted Question 1
If $$x, y$$ and $$z$$ are $${p^{{\text{th}}}},{q^{{\text{th}}}}\,{\text{and }}{r^{{\text{th}}}}$$ terms respectively of an A.P. and also of a G.P., then $${x^{y - z}}{y^{z - x}}{z^{x - y}}$$ is equal to:
If $$a, b, c$$ are in G.P., then the equations $$a{x^2} + 2bx + c = 0$$ and $$d{x^2} + 2ex + f = 0$$ have a common root if $$\frac{d}{a},\frac{e}{b},\frac{f}{c}$$ are in-