Number of possible values of $n$ of an AP The sum of the first $n$ terms of an AP whose first term is a (not necessarily positive) integer and the common difference is $2$, is known to be 153. If $n>1$, then the number of possible values of $n$?
My approach :
$153= \frac{n}{2}[ 2a+ (n-1)\cdot2]$
from this we get;
$(a-1)n+2n^2-153=0$;
So shouldn't be the number of possible values of $n$ be $2$ as $D>0$?
 A: Hint: The fact that the discriminant of $2n^2 + (a-1)n - 153 = 0$ is positive is not really relevant here, since $a$ can have several values. However, the  discriminant gives the number of real solutions, and $n$ is the number of terms, so it must be an integer.
You didn't simplify $153= \frac{n}{2}[ 2a+ (n-1)\cdot2]$ correctly, this should give $153= n[ a+ (n-1)]$, so $(a-1)n + n^2 -  153 = 0$ 
Note that you can write $n^2 + (a-1)n =  153$ as $n(n + (a-1)) =  153$, so $n \mid 153$. 
How many divisors larger than $1$ does $153$ have? Does this give a solution for each divisor?
A: Some initial observations might help. 
As the AP terms sum to $153$ and the common difference is $2$, the AP must consist of sequential odd numbers. Also, the $153= 3\cdot 3\cdot 17$ hence factors are $1,3,9,17,51,153$.
Consider an AP, $AP1$, with positive first term $2r+1$ and with $m$ terms. 
Consider its "mirror AP", $AP2$ with the same last term ($2r+2m-1)$ as $AP1$ but  first term $-(2r-1)$. Number of terms is $2r+m$. The first $r$ terms are negative values of the second $r$ terms so the first $2r$ terms sum to zero. This means $AP2$ has the same sum as $AP1$
Hence, for a given value of $m$ , possible AP lengths $n$ are $m$ and $m+2r$.
Last term is $2r+2m-1$. Sum of AP equals $153$ (given), i.e.
$$\begin{align}
153&=\frac m2\big[(2r+1)+(2r+2m-1)\big]\\
&=m(m+2r)\\
&=1\cdot 153\\
&=3\cdot 51\\
&=9\cdot 17
\end{align}$$
As $n>1$ therefore $\color{red}{n=3, 9, 17, 51, 153}$. 

The possible APs are as follows (terms in blue sum to zero):
$$
\begin{array}
\hline
\hline
&n=m
&n=m+2r\\\\
\hline
m=1: \qquad
&\text{N/A}
&\color{blue}{-151,-149,\cdots 149,151}, 153\\
&&(n=153)\\\\
m=3:\qquad 
&49,51,53
&\color{blue}{-47,-45,\cdots 45,47},49,51, 53\\
&(n=3)
&(n=51)\\\\
m=9:\qquad
&9,11,13,\cdots, 25
&\color{blue}{-7,-5,\cdots 5,7},9,11,13,\cdots, 25\\
&(n=9)
&(n=17)\\
\hline
\end{array}$$
