War between two arithmetic progressions!

From the sum of first $n$ terms of the AP : $S_n=\dfrac{n}2[2a_1+(n-1)d]$
The sum of first $\color{#302B94}n$ terms of $\color{#624F41}{85},90,95,\cdots$ is

$\begin{aligned}\begin{aligned}S_{\color{#302B94}n}=&\ \dfrac{\color{#302B94}n}2[2\color{#624F41}{(85)}+(\color{#302B94}n-1)(5)]\\=&\ \dfrac{n}2(170+5n-5)\\=&\ \dfrac{n}2(165+5n)\\=&\ \color{#3D99F6}{\dfrac{165n+5n^2}2\Rightarrow(1)}\end{aligned}\end{aligned}$

The sum of first $\color{magenta}{3n}$ terms of $\color{#20A900}9,11,13,\cdots$ is

$\begin{aligned}\begin{aligned}S_{\color{magenta}{3n}}=&\ \dfrac{\color{magenta}{3n}}2[2\color{#20A900}{(9)}+(\color{magenta}{3n}-1)(2)]\\=&\ \dfrac{3n}2(18+6n-2)\\=&\ \dfrac{3n}2(16+6n)\\=&\ 3n(8+3n)\\=&\ \color{#D61F06}{24n+9n^2\Rightarrow(2)}\end{aligned}\end{aligned}$

Equation $\color{#3D99F6}{(1)}=$ Equation $\color{#D61F06}{(2)}$ That is,

$\begin{aligned}\begin{aligned}\color{#3D99F6}{\dfrac{165n+5n^2}2}=&\ \color{#D61F06}{24n+9n^2}\\165n+5n^2=&\ 48n+18n^2\\13n^2-117n=&\ 0\\n^2-9n=&\ 0\\n(n-9)=&\ 0\\n=&\ 0,9\end{aligned}\end{aligned}$

But the value of $n$ must be a $\mathbb Z^+$ . Thus, $n=\boxed9$

Given That : => n/2(170+(n-1)5) = 3n/2(18+(3n-1)2) => 165 + 5n = 3(16 + 6n) => 165 - 48 = 18n - 5n => 117 = 13n => n = 9.