How many terms of the series $-12,-9,-6\ldots$ must be taken so that the sum may be $54$?

Question

My attempt: we need to find $n$ given here $a = -12$, $d = 3$, sum $= 54$,

so , $${n\over 2}( 2a + (n-1)d ) = 54$$
$${n\over 2}(2 \cdot (-12) + (n-1)3 ) =54$$
$${n\over 2}(-24 + 3n - 3) = 54$$
$${n\over 2}(-27 + 3n ) = 54$$
$$-27n + 3n^2 = 108$$

I am stuck here.

Answer 1

We can continue from this point:

$$-27n + 3n^2 = 108$$

We want to turn this to this form that has a known solution formula:

$$An^{2}+Bn+C=0$$

Add (-108) to both sides:

$$-27n + 3n^2 -108=0$$

Arrange the terms according to powers of $(n)$:

$$3n^{2}-27n-108=0$$

Divide both sides by $(3)$ to simplify the looks (optional):

$$n^{2}-9n-36=0$$

The general solution(s) for the quadratic equation are:

$$n=\left(\frac{-B+\sqrt{\left(B^2-4AC\right)}}{2A}\right)\:And,\:Or\:n=\left(\frac{-B-\sqrt{\left(B^2-4AC\right)}}{2A}\right)$$

We should get 1 solution only in our case, it can't be negative because its a count.

In our case: $A=1$ and $B=-9$ and $C=-36$

The correct answer is $12$.

The $12$ terms would be: $$-12, -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21$$

Answer 2

It seems the OP is seeking the shortest method to solve this math puzzle. Here is a short-cut.

Take a look at te first 9 terms: $-12, -9, -6, -3, 0, 3, 6, 9,12$. It is clear that all terms cancel so their sum is equal to zero.

Now add a few additional terms ($15, 18, 21$) and you have reached the total of $54$.