Question

If the first and $(2n-1)th$ terms of an A.P., a G.P. and H.P., are equal and their $nth$ terms are $4a,b$, and $c$ respectively, then

• A. $a=b=c$
• B. $a+c=b$
• C. $a>b>c$
• D. $ac-b^2=0$

Answer 1

Answer: (C), (D)

The correct options are
C $a>b>c$
D $ac-b^2=0$

let the first term is $x$

$(2n-1)th$ term is $y$

$y=x+(2n-2)d$ (for AP)

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

$y=x{r}^{2n-2}$ (for GP)

$r=(\frac{y}{x}{)}^{\frac{1}{2n-2}}$
$\frac{1}{y}=\frac{1}{x}+(2n-2)d_1$ (for HP)

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

given $nth$ terms are $a,b,c$

$a=x+(n-1)d$ (for A.P.)

$a=\frac{x+y}{2}$ (after putting value of d)

$b=x{r}^{n-1}$ (for G.P.)

$b=\sqrt{xy}$ (after putting value of r)

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

$c=\frac{2xy}{x+y}$ (after putting value of $d_1$)

finding $ac$

$ac=xy=b^2$ $\frac{b}{c}=\frac{a}{b}$.......(1)

$ac-b^2=0$

finding $a-b$

$a-b=\frac{1}{2}(\sqrt{x}-\sqrt{y}{)}^2$

$a-b\ge 0$

$a\ge b$

from Eq (1)

$b\ge c$

$a\ge b\ge c$