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

Let $A$ be the first term, $D$ be the common dif-ference and $B$ be the $(2n+1)$th term of the $A$

$B=A+2nD\Rightarrow D=\frac{\overline{B}-A}{2n}.$

Also, $a=A+(n+1-1)D=\frac{A+B}{2}.$

Similarly, $b=\sqrt{AB}$ and $c=\frac{2AB}{A+B}.$

Clearly, we have              $b^2=ac$