Let $a_{1}, a_{2}, a_{3}, \dots,a_{n}$ be in ap. given $\sum\limits_{r=1}^\mathbb{10}a_{r}^2 = 1185.$ Find max value of $n $ for $S(n)\ge S(n-1)$

Question

Let $a_{1}, a_{2}, a_{3}, \dots,a_{n}$ be real numbers in an arithmetic progression such that $a_{1}=15$ and $a_{2}$ is an integer. Given $\sum\limits_{r=1}^\mathbb{10}a_{r}^2 = 1185$. If $S(n)=\sum\limits_{r=1}^\mathbb{n}a_{r}$, find the maximum value of $n$ for which $S(n)\ge S(n-1).$

My approach:

I tried to write the sum of every term squared and made a quadratic in $d$ (difference), which is $$285d^2+900d+1065=0,$$ which doesn't have any roots. I might have made a calculation mistake, I have checked multiple times but still be vary. The answer for the maximum value of $n$ is $16.$

EDIT: I made a calculation error, the quadratic formed is $$285d^2+1350d+1065=0,$$

Answer 1

It should be $$285d^2+1350d+1065=0,$$ which gives $d=-1$.

Can you end it now?