Possible number of terms in an Arithmetic Progression

Question

The sum of the first $n$ $(n>1)$ terms of the A.P. is $153$ and the common difference is $2$. If the first term is an integer , then number of possible values of $n$ is

$a)$ $3$

$b)$ $4$

$c)$ $5$

$d)$ $6$

My approach : I used the formula for the first $n$ terms of an A.P. to arrive at the following quadratic equation $n^2 + n(a-1) -153 = 0 $

Next up I realised that since we are talking about the number of terms , thus the possible values which n can take must be whole numbers. That is the discriminant of the above quadratic should yield a whole number in other words

$ (a-1)^2 + 612 = y^2 $ for some y .

However I am stuck at this point , as from here I am unable to figure out the number of such a's ( i.e. the initial terms of an AP ) which will complete the required pythagorean triplet The answer mentioned is $5$

Please let me know , if I am doing a step wrong somewhere . Or If you have a better solution , that will be welcomed too.

Answer 1

To summarize the (extensive) discussion in the comments:

The OP's method is sound and nearly complete. To finish it off we look at the relation $$612=y^2-(a-1)^2=(y+(a-1))(y-(a+1)$$ To solve that (over the integers) we simply need to factor $612=cd$ where the factors must have the same parity. There are three possible such factorings: $$\{18,34\},\;\{2,306\},\:\{6,102\}$$

Each of these gives rise to two possible starting points for our progressions. We get $$a\in \{-151,-47,-7,9,49,153\}$$

We reject the "degenerate" case $a=153$ as that progression just has a single term (and the OP specified $n>1$). Thus we have $5$ solutions.

Answer 2

one of the possible value of $n$ is $3$.

REASON

Let the first term of $AP$ be $a-2$.

$AP: a-2, a, a+2,a+4,a+6,a+8,a+10,\cdots$

Sum of first three terms of $AP= 3a$ which will give $a=51$(integer value).

Now generalising this pattern

we get other possible values of $n$ are $9,17,51,153$

As when we get odd number of terms say, $2k+1$

We can take $AP$ to be $a-2k,a-2(k-1),a-2(k-2),\cdots,a,a+2,a+4,\cdots,a+2(k-1),a+2k\cdots$

(Note: I have mentioned first $2k+1$ terms here)

On adding all these we get $a(2k+1)$ (as common difference will cancel each other due to negative pairity)

Thus $a(2k+1)=153\Rightarrow a=\frac{153}{2k+1}$ ,for integral value of $a, 2k+1$ has to be a positive factor of $153$ which are $5$(excluding 1).

Answer 3

Interesting question (+$1$).

As the common difference is $2$, the series is either one of odd numbers or even numbers only. As the sum is an odd number, it must be a series of odd numbers with an odd number of terms. This narrows it down to $3$ terms or $5$ terms from the choices given.

Since $153\div 3=51$, a quick check shows that $49+51+53=153$ hence the answer is $3.\;\blacksquare$

NB: $153\div 5\approx 30$. The four closest odd numbers to $30$ add up to $120$, and the next one is either $25$ or $35$, none of which would result in $153$.