Question

Let the $n^{th}$ terms of an A.P., G.P. and H.P. be $a,b,c$ respectively. If the first and the $(2n-1{)}^{th}$ terms of the A.P., G.P. and H.P. are equal, then which of the following is/are correct?

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

Answer 1

Answer: (B), (D)

$ac-b^2=0$

Let the first and $(2n-1{)}^{th}$ term be $x$ and $y$ respectively,
We know that,
$1^{st},n^{th},(2n-1{)}^{th}$ terms are in A.P.
$x,a,y$ are in A.P.

$1^{st},n^{th},(2n-1{)}^{th}$ terms are in G.P.
$x,b,y$ are in G.P.

$1^{st},n^{th},(2n-1{)}^{th}$ terms are in H.P.
$x,c,y$ are in H.P.

Now,
$A.M.=a=\frac{x+y}{2}$
$G.M.=b=\sqrt{xy}$
$H.M.=c=\frac{2xy}{x+y}$

Using relation between A.M., G.M. and H.M., we get
$b^2=ac\therefore ac-b^2=0$

Also, using A.M., G.M. and H.M. inequality, we get
$a\ge b\ge c$