# Alycia's question at Yahoo! Answers regarding finding the first term in an arithmetic series

#### MarkFL

Staff member
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

Staff member
Hello Alycia,

I would begin by writing the series as follows:

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

Next, we may use the following:

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

to write:

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

Then, we may use the following:

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

where $a$ is a constant, to write:

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

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

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

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

to write:

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

Solving for $a_1$, we find:

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

Finally, plugging in the given data, we find:

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