just show that in this sequence there exists tuples of the form( k^2 +3k , k^2 + 6k, k^2 + 9k + 18) such that they are in h.p and the second one automatically follows for all positive integer k

Can you please provide the motivation for choosing such tuples?

Denote the set $\{1,4,7,10,\ldots\}$ by $S.$ Suppose $n\in S.$ Hence $2(n+1)\in S.$ Also $2n(n+1)\in S.$ Finally $4n\in S.$ We see that $\frac{1}{2(n+1)}+\frac{1}{2n(n+1)}=\frac{(n+1)}{2n(n+1)}=\frac{1}{2n}=\frac{2}{4n}.$ Hence $\{a_1,a_2,a_3\}\longrightarrow\{2(n+1),4n,2n(n+1)\}$ is a possible triple. We see that for two such distinct triples $\{a_1,a_2,a_3\}\longrightarrow\{2(n+1),4n,2n(n+1)\}$ and $\{b_1,b_2,b_3\}\longrightarrow\{2(m+1),4m,2m(m+1)\}$ with $n\ne m,$ we have $\frac{a_1}{b_1}=\frac{n+1}{m+1}$ and $\frac{a_2}{b_2}=\frac{n}{m}.$ Clearly we see that $\frac{a_1}{b_1}\ne\frac{a_2}{b_2}.$ Hence as $n$ varies in $S,$ we will get infinitely many triples satisfying the required condition.

Good good solution man Bashy99 wrote: Denote the set $\{1,4,7,10,\ldots\}$ by $S.$ Suppose $n\in S.$ Hence $2(n+1)\in S.$ Also $2n(n+1)\in S.$ Finally $4n\in S.$ We see that $\frac{1}{2(n+1)}+\frac{1}{2n(n+1)}=\frac{(n+1)}{2n(n+1)}=\frac{1}{2n}=\frac{2}{4n}.$ Hence $\{a_1,a_2,a_3\}\longrightarrow\{2(n+1),4n,2n(n+1)\}$ is a possible triple. We see that for two such distinct triples $\{a_1,a_2,a_3\}\longrightarrow\{2(n+1),4n,2n(n+1)\}$ and $\{b_1,b_2,b_3\}\longrightarrow\{2(m+1),4m,2m(m+1)\}$ with $n\ne m,$ we have $\frac{a_1}{b_1}=\frac{n+1}{m+1}$ and $\frac{a_2}{b_2}=\frac{n}{m}.$ Clearly we see that $\frac{a_1}{b_1}\ne\frac{a_2}{b_2}.$ Hence as $n$ varies in $S,$ we will get infinitely many triples satisfying the required condition.

can someone please give the motivation of how to reach to such sequences

For any $\{a_1, a_2, a_3\}\in \{1,4,7,... \}=S$ such that $\{a_1, a_2, a_3\}$ in HP we must have $a_3>a_2>a_1$ and $a_2(a_1+a_3)=2a_1*a_3\implies \frac{a_1}{a_3}=\frac{a_1-a_2}{a_2-a_3}=\frac{1}{k}$ So $a_3=a_1*k, 2a_3=a_2(k+1)$ here $k\equiv 1\mod 3$ so $k\in S$ Now we can rewrite the above equation as $2k*a_1=a_2(k+1)\implies a_2=2*k*k_1, a_1=k_1*(k+1), a_3=k_1*k*(k+1)$ for $k_1\equiv 2\mod 3$ Hence there exist infinitely many sequence $\{a_1, a_2, a_3\}$ and clearly we can obtain another such sequences$\{b_1,b_2,b_3\}$ when $\{k,k_1\}$ changes, for which we have $\frac{a_1}{b_1}\ne\frac{a_2}{b_2}$