# Find the first 10 terms of the sequence. {eq}a_1 = x {/eq}, {eq}d = 2x {/eq}

## Question:

Find the first 10 terms of the sequence.

{eq}a_1 = x {/eq}, {eq}d = 2x {/eq}

## Arithmetic Sequence

An arithmetic sequence is a pattern of numbers where the difference between the current term and the preceding term is always the same. Arithmetic sequences can be used to represent many things in mathematics. For example, you might use them to count the number of coins in a pocket or how many times you've done a task today.

Recall that the {eq}n^{th} {/eq} term of an arithmetic sequence is calculated as:

$$a_n=a_1+(n-1)d$$

Where:

• {eq}a_1 {/eq} is the first term and,
• {eq}d {/eq} is the common difference.

Given that {eq}a_1=x\text{ and } d=2x {/eq}, then the expression for {eq}a_n {/eq} is:

\begin{align} a_n&=x+(n-1)2x\\[0.3cm] &=x+2nx-2x\\[0.3cm] &=2nx-x\\[0.3cm] &=(2n-1)x \end{align}

Using the above expression, we have:

The first term is:

$$\bf \boxed{a_1=x}$$

The second term is:

\begin{align} a_2&=(2\cdot 2-1)x\\[0.3cm] &=(4-1)x\\[0.3cm] &=\bf \boxed{3x} \end{align}

The third term is:

\begin{align} a_3&=(2\cdot 3-1)x\\[0.3cm] &=(6-1)x\\[0.3cm] &=\bf \boxed{5x} \end{align}

The fourth term is:

\begin{align} a_4&=(2\cdot 4-1)x\\[0.3cm] &=(8-1)x\\[0.3cm] &=\bf \boxed{7x} \end{align}

The fifth term is:

\begin{align} a_5&=(2\cdot 5-1)x\\[0.3cm] &=(10-1)x\\[0.3cm] &=\bf \boxed{9x} \end{align}

The sixth term is:

\begin{align} a_6&=(2\cdot 6-1)x\\[0.3cm] &=(12-1)x\\[0.3cm] &=\bf \boxed{11x} \end{align}

The seventh term is:

\begin{align} a_7&=(2\cdot 7-1)x\\[0.3cm] &=(14-1)x\\[0.3cm] &=\bf \boxed{13x} \end{align}

The eighth term is:

\begin{align} a_8&=(2\cdot 8-1)x\\[0.3cm] &=(16-1)x\\[0.3cm] &=\bf \boxed{15x} \end{align}

The ninth term is:

\begin{align} a_9&=(2\cdot 9-1)x\\[0.3cm] &=(18-1)x\\[0.3cm] &=\bf \boxed{17x} \end{align}

The tenth term is:

\begin{align} a_{10}&=(2\cdot 10-1)x\\[0.3cm] &=(20-1)x\\[0.3cm] &=\bf \boxed{19x} \end{align}