Show that the infinite arithmetic progression $$ has infinitely many 3 term subsequences in harmonic progression

Question

If $a,b,c$ are in hp then $\dfrac{2ac}{a+c}=b$

Let the 3 term subsequence be

$ 3m+1,3n+1,3p+1 $

Substituting in above formula yields

$6pm-3mn-3pn+m+p -2n = 0$

Then what to do further?? QUESTION further asks to prove(not included in title)

that for any two such triples $<a_1,a_2,a_3>$ and $<b_1,b_2,b_3>$ in harmonic progression one has $\frac{a_1}{b_1}\not = \frac{a_2}{b_2}$

Answer 1

For any $m\in\{1,2,3,\dots\}$, let $n=2m$ and $p=6m^2+3m$.

Then $a=3m+1, b=3n+1=6m+1, $ and $c=3p+1=18m^2+9m+1$,

so $ \dfrac{2ac}{a+c}=\dfrac{2(3m+1)(18m^2+9m+1)}{18m^2+12m+2}=\dfrac{ 18m^2+9m+1}{3m+1}=6m+1=b.$

This works for any $n$,

so there are infinitely many $3$-term subsequences $(a,b,c)$ of $1,4,7,\dots$ in harmonic progression.