Number of possible values of $n$ of an AP

Question

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 $D>0$?

Answer 1

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?

Answer 2

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 $\color{red}{n=3, 9, 17, 51, 153}$.

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The possible APs are as follows (terms in blue sum to zero):

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