# Arithmetic Sequence Problems

An arithmetic sequence is a sequence where there is a common difference between any two successive terms.

$$\large \require{AMSsymbols} \require{color} \color{red} u_{n} = u_{1}+(n-1)d$$

where $\require{color} \color{red} u_{1}$ is the first term and $\require{color} \color{red}d$ is the common difference of the arithmetic sequence.

## Example 1

A city studies and found a population of $5000$ in the first year of the study. The population increases by $200$ each year after that.

(a) β Write down a rule for the population in month $n$ of the study.

\begin{align} \displaystyle u_{n} &= 5000+(n-1)\times 200 \\ &= 5000 + 200n-200 \\ u_{n} &= 200n + 4800 \\ \end{align}

(b) β When will the population double in size?

\begin{align} \displaystyle 200n + 4800 &= 10000 \\ 200n &= 5200 \\ n &= 5200 \div 200 \\ &= 26 \\ \end{align}
Thus the population will double in the $26$th month.

## Example 2

For the arithmetic sequence $\{21,x,y,36\}$, find the values of $x$ and $y$.

\begin{align} \displaystyle u_{1} &= 21 \\ u_{2} &= 21 + d = x \cdots (1) \\ u_{3} &= 21 + 2d = y \cdots (2) \\ u_{4} &= 21 + 3d = 36 \cdots (3) \\ 3d &= 36-21 \cdots (3) \\ 3d &= 15 \\ d &= 5 \\ x &= 21 + 5 &\text{substitute } d = 5 \text{ into } (1) \\ &= 26 \\ y &= 21 + 2 \times 5 &\text{substitute } d = 5 \text{ into } (2) \\ &= 21+10 \\ &= 31 \\ \therefore x &= 26 \text{ and } y=31 \\ \end{align}

## Example 3

Find the value of $x$ such that $\{\cdots,x,3x+4,10x-7,\cdots\}$ forms an arithmetic sequence.

An arithmetic sequence is a sequence where there is a common difference between any two successive terms.
\begin{align} \displaystyle (3x+4)-(x) &= (10x-7)-(3x+4) \\ 3x+4-x &= 10x-7 – 3x-4 \\ 2x+4 &= 7x-11 \\ 2x-7x &= -11-4 \\ -5x &= -15 \\ \therefore x &= 3 \end{align}

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