Inspired by Abhiram Rao

$2015 a_{2015} = 2016 a_{2016}\\ 2015(a + 2014d) = 2016(a + 2015d)\\ 2015a + 2015 × 2014d = 2016a + 2015 × 2016d\\ 2015a - 2016a = 2015 × 2016d - 2015 × 2014d\\ -a = 2015d(2016 - 2014)\\ a = -4030d$

Let the nth term in the AP be $0$ . Then $a + (n-1)d = 0\\ \implies a = -(n-1)d\\ \implies -4030d = -(n-1)d\\ 4030 = (n-1)\\ \implies n = \color{#69047E}{\boxed{4031}}$ .

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Generalizing:-
Let $n$ times $a_n$ = $m$ times $a_m$ .
Then $n(a + (n - 1)d) = m(a + (m - 1)d)\\ n(a + dn - d) = m(a + dm - d)\\ an + dn^2 - dn = am + dm^2 - dm\\ an - am + dn^2 - dm^2 - dn + dm = 0\\ a(n - m) + d\left(n^2 - m^2 - n + m\right) = 0\\ a(n - m) + d\left(n^2 - m^2 - (n - m)\right) = 0\\ a(n - m) + d\left(\left(n - m\right)\left(n + m - 1\right)\right) = 0\\ (n - m)\left(a + (n + m - 1)d\right) = 0\\ a + (n + m - 1)d = 0\\ a_{n + m} = 0$

So, if ever, n times nth term is equal to m times mth term of a non-constant arithmetic progression, then the (m + n)th term is $0$ .

$2015a_{2015} = 2016a_{2016}$

$2015a_{2015} = 2016(a_{2015} + d)$ .....................................Let d be the common difference. So obviously $a_{2016}=a_{2015} + d$

$2015a_{2015}=2016a_{2015} + 2016d$

$2016a_{2015} - 2015a_{2015} + 2016d = 0$

$a_{2015} + 2016d = 0$

So clearly when you add common difference 2016 times you will get $a_{2015 + 2016}$

Which is $a_{4031}$ ............. 4031 st term in the series

Therefore the answer is $\boxed{4031}$

$\begin{aligned} 2016a_{2016} & = 2015a_{2015} \\ 2016a_{2016} - 2015a_{2015} & = 0 \\ a_{2016} + 2015(\color{#3D99F6}{a_{2016} - a_{2015}}) & = 0 \\ a_{2016} + 2015\color{#3D99F6}{d} & = 0 \quad \quad \small \color{#3D99F6}{d = \text{common difference}} \\ \implies \boxed{a_{\color{#3D99F6}{4031}}} & = 0 \quad \quad \small \color{#3D99F6}{2016+2015=4031} \end{aligned}$

The $\boxed{4031}$ th term is equal to 0.

I have done tuhis by the same. Thanks for the solution.

Let $c$ be the common difference and $a_n$ the term which is zero. Then $2015a_{2015} = 2016(a_{2015}+c)$ and $a_{2015} + 2016c = 0 = a_n$ . Given that in an arithmetic progression the $n^{th}$ therm can be expressed as $a_n = a_k + (n-k)c$ , we have that $2016 = n-2015$ and therefore $n = 4031$ .