Question

If $a,b,c,d$ are distinct positive numbers which are in A.P. with positive common difference, then which of the following is/are correct?

• A. $\frac{1}{a}+\frac{1}{d}>\frac{1}{b}+\frac{1}{c}$
• B. $\frac{1}{a}+\frac{1}{d}<\frac{1}{b}+\frac{1}{c}$
• C. $\frac{1}{b}+\frac{1}{c}>\frac{4}{a+d}$
• D. $\frac{1}{b}+\frac{1}{c}<\frac{4}{a+d}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{1}{a}+\frac{1}{d}>\frac{1}{b}+\frac{1}{c}$
C $\frac{1}{b}+\frac{1}{c}>\frac{4}{a+d}$

Let the common difference be $D$

Now,
$\frac{\frac{1}{a}+\frac{1}{d}}{\frac{1}{b}+\frac{1}{c}}=\frac{bc(a+d)}{ad(b+c)}=\frac{bc}{ad}(\because a+d=b+c)=\frac{b(b+D)}{(b-D)(b+2D)}=\frac{(b^2+Db)}{(b^2+bD)-2D^2}>1(\because D\ne 0)\therefore \frac{1}{a}+\frac{1}{d}>\frac{1}{b}+\frac{1}{c}$

and
$(\frac{1}{b}+\frac{1}{c})(a+d)=\left[\frac{(b+c)(a+d)}{bc}\right]=\frac{(b+c{)}^2}{bc}=\frac{b^2+c^2+2bc}{bc}=\frac{b}{c}+\frac{c}{b}+2$
We know that, when $x$ is positive real number, then $x+\frac{1}{x}\ge 2$, so
$\frac{b}{c}+\frac{c}{b}\ge 2$
As $b$ and $c$, are distinct positive numbers, so
$\Rightarrow \frac{b}{c}+\frac{c}{b}>2$

$\therefore \frac{1}{b}+\frac{1}{c}>\frac{4}{a+d}$