Question

If the first and the $(2n-1)th$ terms of an AP, a GP and an HP are equal and positive and their $nth$ terms are $a,b$ and $c$ 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)

The correct options are
B $a\ge b\ge c$
D $ac-b^2=0$

Let $x$ be the $1st$ term and $y$ be the $(2n-1)th$ term of A.P, G.P and H.P. whose $nth$ terms are $a$, $b$ and $c$ respectively.
For A.P, $y=x+(2n-2)d$
$\Rightarrow d=\frac{(y-x)}{2(n-1)}$
$\therefore a=x+(n-1)d=\frac{(x+y)}{2}$
For G.P, $y=x{r}^{2n-2}\Rightarrow r={\left(\frac{y}{x}\right)}^{\frac{1}{2n-2}}$
$\therefore b=x{r}^{n-1}=\sqrt{xy}$
For H.P, $\frac{1}{y}=\frac{1}{x}+(2n-2)d_1$
$\Rightarrow d_1=\frac{(x-y)}{2xy(n-1)}$
$\therefore \frac{1}{c}=\frac{1}{x}+(n-1)d_1=\frac{(x+y)}{2xy}$
Or $c=\frac{2xy}{x+y}$
Now $a-b=\frac{1}{2}(x+y)-\sqrt{xy}=\frac{1}{2}{(\sqrt{x}-\sqrt{y})}^2\ge 0\Rightarrow a\ge b$
Now $ac=b^2\therefore \frac{b}{c}=\frac{a}{b}\ge 1\Rightarrow b\ge c$
Hence, $a\ge b\ge c$ and also $ac-b^2=0$
Thus options (B) and (D) are correct.