Question

The sum of three numbers in H.P is $37$ and the sum of their reciprocals $\frac{1}{4}$. Find the smallest of those three numbers.

Answer 1

Three numbers in H.P are taken as,
$\frac{1}{a-d},\frac{1}{a},\frac{1}{a+d}$
then, $\frac{1}{a-d}+\frac{1}{a}+\frac{1}{a+d}=37\cdots \left(1\right)$
and $a-d+a+a+d=\frac{1}{4}$
$\Rightarrow a=\frac{1}{12}$
from (1),

$\frac{12}{1-12d}+12+\frac{12}{1+12d}=37$
$\Rightarrow \frac{12}{1-12d}+\frac{12}{1+12d}=25$
$\Rightarrow \frac{24}{1-144d^2}=25$
$\Rightarrow 1-144d^2=\frac{24}{25}$
$\Rightarrow d^2=\frac{1}{25\times 144}$
$\therefore d=\pm \frac{1}{60}$
$\therefore a-d,a,a+d$ are $\frac{1}{15},\frac{1}{12},\frac{1}{10}or\frac{1}{10},\frac{1}{12},\frac{1}{15}$
Hence three numbers in H.P are

$15,12,10or10,12,15$