Find the first 10 terms of the sequence.

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

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:

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