The sum of 3 numbers in H.P. is 37 and sum of their reciprocals is $\frac{1}{4}$, then

$\frac{1}{a-d}+\frac{1}{a}+\frac{1}{a+d}=37$

$\Rightarrow a-d+a+a+d=\frac{1}{4}$

$\Rightarrow 3a=\frac{1}{4}$

$\Rightarrow a=\frac{1}{12}$

$\Rightarrow \frac{a(a+d)+(a-d)(a+d)+(a-d)a}{(a-d)a(a+d)}=37$

$\Rightarrow \frac{a^2+ad+a^2-d^2+a^2-da}{a(a^2-d^2)}=37$

$\Rightarrow \frac{3a^2-d^2}{a(a^2-d^2)}=37$

$\Rightarrow \frac{3}{144}-d^2=37\times \frac{1}{12}\left(\frac{1}{144}-d^2\right)$

$\Rightarrow 3-144d^2=\frac{37}{12}(1-144d^2)$

$\Rightarrow \frac{37}{12}\times 144d^2-144d^2=\frac{37}{12}-3$

$\Rightarrow 144d^2\left(\frac{37}{12}-1\right)=\frac{1}{12}$

$\Rightarrow 144d^2.\frac{25}{12}=\frac{1}{12}$

$\Rightarrow d^2=\frac{1}{144\times 25}$

$\Rightarrow d=\frac{1}{12\times 5}=\frac{1}{60}$

$\frac{1}{a-d}=\frac{1}{\frac{1}{12}-\frac{1}{60}}$

$=\frac{1}{\frac{4}{60}}=\frac{1}{\frac{1}{15}}=15$

$\frac{1}{a}=12$

$\frac{1}{a+d}=\frac{1}{\frac{1}{12}+\frac{1}{60}}=\frac{1}{\frac{6}{60}}=10$