If the first and the $${\left( {2n - 1} \right)^{th}}$$ terms of an AP, a GP and an HP are equal and their $$n^{th}$$ terms are $$a, b$$ and $$c$$ respectively then
A.
$$a = b = c$$
B.
$$a \geqslant b \geqslant c$$
C.
$$a + c = b$$
D.
$$ac - {b^2} = 0$$
Answer :
$$ac - {b^2} = 0$$
Solution :
$${n^{th}}$$ term is the middle term in each case. So $$a, b, c$$ are the AM, GM, HM respectively of the same two numbers. For any two numbers AM, GM, HM are in GP.
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-