Find the next three terms in each sequence.

$5,-10,15,-20,\mathrm{25......}$

In anArithmetic Sequencethe difference between one term and the next is a constant.

The nthterm of the Arithmetic sequence is given by the formula

$$\begin{gathered} a_n=a+\left(n-1\right)d \\ d=a_2-a_1 \end{gathered}$$

Where,$a_1=a=$ first term and $d=$common difference.

The above sequence consists of alternating positive and negative terms and so on the pattern will follow for the rest of the terms.

That is, even terms have negative sign.

Rewrite the above sequence by removing the negative sign

$5,10,15,20,25,\mathrm{....}$ (1)

Notice now the above sequence is in arithmetic progression.

The common difference is calculated as follows.

$$\begin{gathered} d=a_2-a_1 \\ d=10-5 \\ d=5 \end{gathered}$$

The nth term of the arithmetic sequence is given as follows.

$\text{Here,\hspace{0.17em}\hspace{0.17em}first\hspace{0.17em}\hspace{0.17em}term\hspace{0.17em}}(a\text{)=5\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}d\text{=5}$

$$\begin{gathered} a_n=5+\left(n-1\right)5 \\ =5+5n-5 \\ {a}_n=5n\dots \left(2\right) \end{gathered}$$

Using equation (2) find the next three terms of the given sequence.

$\begin{array}{c}{a}_6=5\left(6\right)=30\\ a_7=5\left(7\right)=35\\ a_8=5\left(8\right)=40\end{array}$

Give negative sign to even terms,

$\begin{array}{c}{a}_6=-30\\ a_7=35\\ a_8=-40\end{array}$

Therefore, the next three terms are $-30,35\mathrm{and}-40$.