Question

Find the first four terms of the sequence given by the following.an = 63 +(n-1)5, n=1, 2, 3...

Answer 1

We are required to find the first 4 terms of the sequence an = 63 +(n-1)5

The nth term of the sequence is given by the formula

an = 63 +(n-1)5

When n = 1

`\begin{gathered} a_1=63+(1-1)5\text{ = 63 + (0)5= 63+0} \\ a_1=63 \end{gathered}`

The first term is 63

when n=2

`\begin{gathered} a_2=63+(2-1)5\text{ = 63 + (1)5= 63+}5 \\ a_2=68 \end{gathered}`

The second term is 68

When n=3

`\begin{gathered} a_3=63+(3-1)5\text{ = 63 + (2)5= 63+}10 \\ a_3=73 \end{gathered}`

The third term is 73

When n = 4

`\begin{gathered} a_4=63+(4-1)5\text{ = 63 + (3)5= 63+}15 \\ a_4=78 \end{gathered}`

The fouth term is 78

Hence, the first 4 terms of the sequence are 63, 68, 73, 78