Question

Find the next three terms of each arithmetic sequence.

$6,11,16,21,\mathrm{...}$.

Short answer

The next three terms of the given arithmetic sequence are 26, 31 and 36.

Step-by-step solution

Step 1. Find the common difference (d) of the given arithmetic sequence.

The given arithmetic sequence is:$6,11,16,21,\mathrm{...}$

Therefore, it can be written that:

$$\begin{gathered} a_1=6 \\ {a}_2=11 \\ {a}_3=16 \\ {a}_4=21 \end{gathered}$$

The common difference (d) of the arithmetic sequence is the difference between any term and the term prior to that term.

Therefore,

$$\begin{gathered} d=a_2-a_1 \\ =11-6 \\ =5 \end{gathered}$$

Therefore, the common difference (d) of the given arithmetic sequence is 5.

Step 2. Find the next three terms of the given arithmetic sequence.

The nth term of the arithmetic sequence $\left(a_n\right)$is given by:

$a_n=a+\left(n-1\right)d$

Where the first is term of the arithmetic sequence and d is the common difference of the arithmetic sequence.

The next three terms of the given arithmetic sequence are $a_5,a_6$and $a_7$.

The given arithmetic has first term as 6 and common difference as 5.

Therefore,$a=6$ and $d=5$.

Therefore, it is obtained that:

$$\begin{gathered} a_5=6+\left(5-1\right)\left(5\right) \\ =6+4\left(5\right) \\ =6+20 \\ =26 \end{gathered}$$

$$\begin{gathered} a_6=6+\left(6-1\right)\left(5\right) \\ =6+5\left(5\right) \\ =6+25 \\ =31 \end{gathered}$$

$$\begin{gathered} a_7=6+\left(7-1\right)\left(5\right) \\ =6+6\left(5\right) \\ =6+30 \\ =36 \end{gathered}$$

Therefore, the next three terms of the given arithmetic sequence are 26, 31 and 36.