How many terms of the arithmetic sequence -12, -3, 6, ... must be added to arrive at a sum of 363?

Question

So the common difference is +9.

I'm getting confused how to use the formula to figure this out. I wrote a C++ program to help me solve it so I got the answer but I still need to know how to do it mathematically.

The formula to solve for the total is:

$S_n = \frac{n}{2}(a_1 + a_n) $

Here's what I can plug in:

$363 = \frac{n}{2}(-12 + a_n)$

but I can't solve it if I don't know what $a_n$ is equal to. To solve for $a_n$, I use the formula.

$a_n = a_1 + (n - 1)d$

Here's what I can plug in.

$a_n = -12 + (n - 1)9$

I'm confused... How can I solve either one if I need the variable that the other one helps me solve to use it for the other formula? It's like a paradox. I need to solve n in the first equation to solve for $a_n$, but I also need $a_n$ to solve for n.

Answer 1

Substitute the expresion for $a_n$ in the formula of $S_n$:

$$363=\frac n2\big(-12+[-12+(n-1)9]\big)$$

Answer 2

You can use Gauss's formula. It tells you the sum of an arithmetic sequence $a_1+a_2+\dots a_n$ is $\frac{(a_1+a_n)n}{2}$.

In this case we get that the sum of the first $n$ terms is $\frac{(-12+(-12+9(n-1)))n}{2}=\frac{(9n-9-24)n}{2}=\frac{(9n-33)n}{2}$. This is because the first term is $-12$ and the $n$'th term is $-12+(n-1)9$.

We want $\frac{(9n-33)n}{2}=363$ so $(9n-33)n=726\iff 9n^2-33n-726=0$.

This quadratic has roots $\frac{-22}{3}$ and $11$, so $11$ is your answer.

Answer 3

$S_n={n\over 2}(2a_1+(n-1)d)$

from the question,

$a_1= -12$ (i.e first term) and the common difference $d= (-3)-(-12) = 9$ and as sum to get is $363$ that will be $S_n$.

$S_n= 363$. Substitute every term in the above formula, $$363= {n\over 2}[2(-12)+(n-1)(9)]$$ $$363= {n\over 2}[-24+9n-9]$$ by solving the equation $n= {-22\over 3}$ and $11$. hence $n$-th term will be $11$. :)