Finding the sum of the first term and common difference when given sum of first 5 and sum of first 10

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

$\begin{array}{c| cccccccccccccc} \text{index} & 1 & 2 & 3 & 4 & 5\\ \text{term} & a & a+d & a+2d & a+3d & a+4d \\ \text{sum} & a & 2a+d & 3a+3d & 4a+6d & \color{red}{5a+10d=110} \\ \text{formula} &&&&& 5\dfrac{(a)+(a+4d)}{2} \end{array}$

$\begin{array}{c| cccccccccccccc} \text{index} & 6 & 7 & 8 & 9 & 10\\ \text{term} & a+ 5d & a+ 6d & a+ 7d & a+ 8d & a+9d \\ \text{sum} & 6a+15d & 7a+21d & 8a+28d & 9a+36d & \color{red}{10a+45d=320}\\ \text{formula} &&&&& 10\dfrac{(a)+(a+9d)}{2} \end{array}$

\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}