Question

The sum of three numbers in H.P. is $26$ and sum of their reciprocals is $3/8$, Find the largest of the number divided by smallest one ?

Answer 1

Any three numbers in H.P. be taken as $\frac{1}{a-d},\frac{1}{a},\frac{1}{a+d}$ as their reciprocals are in A.P.
Given $\frac{1}{a-d}+\frac{1}{a}+\frac{1}{a+d}=26$...(1)

Also $a-d+a+a+d=\frac{3}{8}a=\frac{1}{8}$

From (1), $\frac{a(a+d)+(a+d)(a-d)+a(a-d)}{a(a^2-d^2)}=26$

$3a^2-d^2=26a(a^2-d^2)=\frac{13}{4}(a^2-d^2)$ Using (1)

$a^2=9d^2$

$d=\pm \frac{a}{3}=\pm \frac{1}{24}$

$Numbersare12,8,6or6,8,12$.

So the Ratio is $2$.