Arithmetic Sequence Problems

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)   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)   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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