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

Let 1^st term $=x;\left(2n-1\right)$ th term $=y$

In A.P; $a_n=x+\left(n-1\right)d$

$a_{2n-1}=x+\left(2n-2\right)d$

$\Rightarrow y=x+2\left(n-1\right)d$

$\Rightarrow d=\frac{y-x}{2\left(n-1\right)}$

$a_n=a$

$\Rightarrow x+\left(n-1\right)d=a$

$\Rightarrow x+\left(n-1\right)\frac{\left(y-x\right)}{2\left(n-1\right)}=a$

$a=\frac{x+y}{2}$

In G.P.,

$a_{2n-1}=x.r^{2n-2}$

$y=x.r^{2\left(n-1\right)}$

$r={\left(\frac{y}{x}\right)}^{\frac{1}{2\left(n-1\right)}}$

$b=a_n$

$\Rightarrow b=x.r^{n-1}$

$\Rightarrow b=x{\left[{\left(\frac{y}{x}\right)}^{\frac{1}{2\left(n-1\right)}}\right]}^{n-1}$

$b=x.{\left(\frac{y}{x}\right)}^{\frac{1}{2}}$

$b=x^{\frac{1}{2}}.y^{\frac{1}{2}}$

$b=\sqrt{xy}$

In H.P.,

$\frac{1}{y}=\frac{1}{x}+\left(2n-2\right)d_1$

$\frac{1}{y}-\frac{1}{x}=2\left(n-1\right)d_1$

$\frac{x-y}{xy}=2\left(n-1\right)d_1$

$\Rightarrow d_1=\frac{x-y}{2\left(n-1\right)xy}$

$\frac{1}{c}=\frac{1}{x}+\left(n-1\right)d_1$

$\Rightarrow \frac{1}{c}=\frac{1}{x}+\frac{\left(n-1\right)\left(x-y\right)}{2\left(n-1\right)xy}$

$\Rightarrow \frac{1}{c}=\frac{2y+\left(x-y\right)}{2xy}$

$\Rightarrow \frac{1}{c}=\frac{x+y}{2xy}$

$\Rightarrow c=\frac{2xy}{x+y}$

$\therefore a,y,c$  are the A.M, G.M, H.M of $x\text{and}y$

We know that $\text{A}\text{.M}\ge \text{G}\text{.M}\ge \text{H}\text{.M}$  and $G^2=\text{A}\text{.H}$

$\Rightarrow a\ge b\ge c$  and $b^2=ac$