Question

If the first and $(2n-1)$ th terms of an A.P, a G.P and a H.P are equal and their $n$th terms are $a,bandc$ respectively, then

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

Answer 1

Answer: (C), (D)

The correct options are
C $a\ge b\ge c$
D $ac=b^2$

let $x$ be the first term and $y$ the $(2n-1)th$ of A.P., G.P., and H.P. whose $nth$ terms are $a,b,c$ respectively.
For A.P; $y=x+(2n-2)d$
$\Rightarrow d=\frac{(y-x)}{2(n-1)}$
Therefore, $a=x+(n-1)d$
$\Rightarrow a=\frac{1}{2}(x+y)$
For G.P; $y=x{r}^{2n-2}$
$\Rightarrow r=(\frac{y}{x}{)}^{\frac{1}{2n-2}}$
Therefore, $b=x{r}^{n-1}$
$\Rightarrow b=\sqrt{xy}$
For H.P; $\frac{1}{y}=\frac{1}{x}+(2n-2)d_1$
$\Rightarrow \frac{1}{y}-\frac{1}{x}=(2n-2)d_1$
$\Rightarrow \frac{(x-y)}{xy}=(2n-2)d_1$
$\Rightarrow d_1=\frac{(x-y)}{2xy(n-1)}$
Therefore, $\Rightarrow \frac{1}{c}=\frac{1}{x}+(n-1)d_1$
$\Rightarrow \frac{1}{c}=\frac{(x+y)}{2xy}$
$\Rightarrow c=\frac{2xy}{(x+y)}$
Now, $a-b=\frac{1}{2}(x+y)-\sqrt{xy}$
$=\frac{1}{2}(\sqrt{x}-\sqrt{y}{)}^2\ge 0$
$\Rightarrow a\ge b$
Consider, $ac=\frac{1}{2}(x+y)\frac{2xy}{(x+y)}$
$\Rightarrow ac=xy$
$\Rightarrow ac=b^2$ $\because b=\sqrt{xy}$
$\Rightarrow \frac{b}{c}=\frac{a}{b}\ge 1$ $\because a\ge b$
$\Rightarrow b\ge c$
Hence, $a\ge b\ge c$