Question

If the first and ${(2n-1)}^{th}$ term of a A.P., G.P. and H.P. are equal and their $n^{th}$ terms are $a,b,c$ respectively, then

• A. $a=b=c$
• B. $a\ge b\ge c$
• C. $a+c=b$
• D. $ac-b^2=0$

Answer 1

Answer: (B), (D)

$ac-b^2=0$

Explanation for the correct options:

Given that first and ${(2n-1)}^{th}$ term of a A.P., G.P. and H.P. are equal, say $x$ and $y$ respectively.

We know that the mean of the first and ${(2n-1)}^{th}$ term is $\frac{1+\left(2n-1\right)}{2}=n^{th}$term i.e. $t_n$

Given that $n^{th}$ terms of A.P., G.P. and H.P. are $a,b,c$ respectively.

$\therefore$ Arithmetic mean A.M. $=\frac{x+y}{2}=a$

Geometric mean G.M.$=\sqrt{xy}=b$

Harmonic mean H.M. $=\frac{2xy}{x+y}=c$

We know that A.M$\ge$G.M.$\ge$H.M$\Rightarrow a\ge b\ge c$

We know that (A.M.$\times$H.M.)^1/2$=$G.M.

$\begin{array}{crcl}\Rightarrow & \sqrt{ac}& =& b\\ \Rightarrow & ac& =& b^2\\ \Rightarrow & ac-b^2& =& 0\end{array}$

Hence, the correct options are option (B) and (D) i.e. $a\ge b\ge c$ and $ac-b^2=0$.