Finding the sum of the first term and common difference when given sum of first 5 and sum of first 10 the sum of the first 5 terms of an arithmetic series is 110 and the sum of the first 10 terms is 320. How do i go about finding the first term and common difference.
Sn  = n/2 [2a+d(n−1)] is the equation for working out the sum of an arithmetic series, but how can i rearrange to find for the first term and common difference. I belive it would be using simulatenous equations.
 A: Sn = n/2 (2a + (n-1) d)
110 = 5/2 (2a+(5–1)d) (Eq. 1)
320=10/2(2a+(10–1)d (Eq.2)
110=2.5(2a+5d-1d) - 110=2.5(2a+4d) (Eq.3)
320=5(2a+10d-1d) - 320=5(2a+9d) (Eq.4)
64=(2a+9d) - Divided both sides 5 from equation 4 (eq.5)
44=2a+4d - Divided both sides 2.5 from equation 3 (eq.6)
20=5d  - Simulataneous Equations - just minus it through. 64–44 is 20. 2a-2a is 0 and  9d-4d is 5d
4=d -20/5 is 4 
Sub d into equation (5)
64 = 2a+9(4)
64=2a+36
28=2a
14=a
Therefore the first term would be 14 and the common difference would be 4
A: Let first term $a_1$ be A (for ease of writing):
$$s_n=\frac{n}{2}[2A+(n-1)d]$$
Since $S_5=110$
$$s_5=110=\frac{5}{2}[2A+(5-1)d]$$
$$44=2A+4d \tag1$$
Since $S_{10}=320$
$$s_{10}=320=\frac{10}{2}[2A+(10-1)d]$$
$$64=2A+9d\tag2$$
Solving (1) and (2), We get:
$$d=4$$
Using either (1) or (2), say (1), we substitute $d=4$ to get $A$:
$$44=2A+4(4)$$
Solving for $A$, we get $$A=14$$
The series is:
$$14, 18, 22, 26, 30, 34, 38, 42, 46, 50, ...$$
Note: Arithmatic Progression-Wiki has a good list of formulae related to this subject.
A: $\begin{array}{c| cccccccccccccc}
\text{index} & 1 & 2    & 3     &  4    & 5\\
\text{term}  & a & a+d  &  a+2d &  a+3d & a+4d \\
\text{sum}   & a & 2a+d & 3a+3d & 4a+6d & \color{red}{5a+10d=110} \\
\text{formula} &&&&& 5\dfrac{(a)+(a+4d)}{2}
\end{array}$
$\begin{array}{c| cccccccccccccc}
\text{index} & 6      & 7      & 8      &  9     & 10\\
\text{term}  &  a+ 5d &  a+ 6d &  a+ 7d &  a+ 8d & a+9d \\
\text{sum}   & 6a+15d & 7a+21d & 8a+28d & 9a+36d & \color{red}{10a+45d=320}\\
\text{formula} &&&&& 10\dfrac{(a)+(a+9d)}{2}
\end{array}$
\begin{align}
    5a+10d &=110 \\
   10a+45d &=320 \\
\hline
    10a+20d &=220 \\
   10a+45d &=320 \\
\hline
   25d &= 100 \\
\hline
   d &= 4 \\
   a &= 14
\end{align}
