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

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To find the first term of an arithmetic series when given the number of terms (n), common difference (d), and sum (Sn), the formula a1 = (2Sn - n(n-1)d) / (2n) can be used. In the example provided, with d = 6, Sn = 574, and n = 14, substituting these values into the formula yields a1 = 2. The calculations involve applying known summation results for series. This method effectively determines the first term in the series. Understanding this formula simplifies the process of finding the first term in similar arithmetic series problems.
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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.
 
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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:

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