Question

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

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

Answer 1

Answer: (B), (D)

$a\ge b\ge c$
$ac-b^2=0$

Let the first term be $1$

$a=1+n{d}_1$

$b=r^n$

$c=\frac{1}{1+n{d}_2}$

$1+2n{d}_1=r^{2n}$$=\frac{1}{2n{d}_2}$

$1+2n{d}_1+2n{d}_2+4n^2{d}_1{d}_1=1$

Or $d_1+d_2+2n{d}_1{d}_1=0$

$n=-\frac{1}{2}(\frac{1}{d_1}+\frac{1}{d_2})$

Now $b^2=r^{2n}$

$ac=\frac{1+n{d}_1}{1+n{d}_2}$

$ac=\frac{1-\frac{1}{2}(\frac{1}{d_1}+\frac{1}{d_2})d_1}{1-\frac{1}{2}(\frac{1}{d_1}+\frac{1}{d_2})d_2}$

$ac=\frac{1-\frac{d_1+d_2}{2d_2}}{1-\frac{1}{2}\frac{d_1+d_2}{d_1}}$

$=-\frac{d_1}{d_2}$

$1+2n{d}_1=1+d_1(-\frac{1}{d_1}-\frac{1}{d_2})$

$=1-\frac{d_1+d_2}{d_2}$

$ac=-\frac{d_1}{d_2}=r^{2n}=b^2$

Thus D is correct

$a=1+(-\frac{1}{2})(\frac{1}{d_1}+\frac{1}{d_2})d_1$

$a=1-\frac{1}{2}\left(\frac{d_1+d_2}{d_2}\right)$

$a=\frac{d_2-d_2}{{2d}_2}$

$c=\frac{-\frac{d_1}{d_2}}{\frac{d_2-d_1}{{2d}_2}}$

$b=\sqrt{-\frac{d_1}{d_2}}$

Now $a\ge c$ because ${(d_2-d_1)}^2=-4d_2{d}_1$

Or ${(d_2+d_1)}^2\ge 0$

And $b\ge c$ because $-\frac{d_1}{d_2}\ge \frac{4d_1}{{(d_2-d_1)}^2}$

${(d_2+d_1)}^2\ge 0$

Thus B is also correct