MHB Find First Term in Arithmetic Series | Alycia's Question

[MarkFL]

Here is the question:

How to find the first term in an arithmetic series?

I missed a day of math so now I have to catch up. How do you find the first term of an arithmetic series when given the number of terms (n), the common difference (d), and the sum (Sn)? The solution is probably obvious and I'm just making it more complicated than it actually is.
Here's a question from the book: Determine the value of the first term (t1) for each arithmetic series described.
d = 6, Sn = 574, n = 14
I know that the answer is 2 (I checked the back of the book), but I can't figure out how to get that answer.
Thanks for anyone her helps. :)

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Alycia,

I would begin by writing the series as follows:

$$S_n=\sum_{k=0}^{n-1}\left(a_1+k\cdot d \right)$$

Next, we may use the following:

$$\sum_{k=k_i}^{k_f}\left(f(k)\pm g(k) \right)=\sum_{k=k_i}^{k_f}\left(f(k) \right)\pm\sum_{k=k_i}^{k_f}\left(g(k) \right)$$

to write:

$$S_n=\sum_{k=0}^{n-1}\left(a_1 \right)+\sum_{k=0}^{n-1}\left(k\cdot d \right)$$

Then, we may use the following:

$$\sum_{k=k_i}^{k_f}\left(a\cdot f(k) \right)=a\cdot\sum_{k=k_i}^{k_f}\left(f(k) \right)$$

where $a$ is a constant, to write:

$$S_n=a_1\cdot\sum_{k=0}^{n-1}\left(1 \right)+d\cdot\sum_{k=0}^{n-1}\left(k \right)$$

Next, we may utilize the following well-known summation results:

$$\sum_{k=1}^n(1)=n$$

$$\sum_{k=1}^n(k)=\frac{n(n+1)}{2}$$

to write:

$$S_n=a_1n+d\frac{n(n-1)}{2}$$

Solving for $a_1$, we find:

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

Finally, plugging in the given data, we find:

$$a_1=\frac{2\cdot574-14\cdot13\cdot6}{2\cdot14}=2$$