Question

How do you write the nth term rule for the arithmetic sequence with `a_7=34` and `a_18=122`?

Answer 1

`n^(th)` term of the arithmetic sequence is `8n-22`.

Explanation:

`n^(th)` term of an arithmetic sequence whose first term is `a_1` and common difference is `d` is `a_1+(n-1)d`.

Hence `a_7=a_1+(7-1)xxd=34` i.e. `a_1+6d=34`

and `a_18=a_1+(18-1)xxd=122` i.e. `a_1+17d=122`

Subtracting firt equation from second equation, we get

`11d=122-34=88` or `d=88/11=8`

Hence `a_1+6xx8=34` or `a_1=34-48=-14`

Hence `n^(th)` term of the arithmetic sequence is `-14+(n-1)xx8` or `-14+8n-8=8n-22`.

Answer 2

`color(blue)(a_n=8n-22)`

Explanation:

The given data are

`a_7=34` and `a_18=122`

We can set up 2 equations

`a_n=a_1+(n-1)*d`

`a_7=a_1+(7-1)*d`

`34=a_1+6*d" "`first equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`a_n=a_1+(n-1)*d`

`a_18=a_1+(18-1)*d`

`122=a_1+17*d" "`second equation

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

By method of elimination using subtraction, let us use first and second equations

`34=a_1+6*d" "`first equation

`122=a_1+17*d" "`second equation

By subtraction, we have the result

`88=0+11d`

`d=88/11=8`

Solving now for `a_1` using the first equation and `d=8`

`34=a_1+6*d" "`first equation

`34=a_1+6*8" "`

`34=a_1+48`

`a_1=-14`

We can write the `nth` term rule now

#a_n=-14+8*(n-1)

`a_n=-14-8+8n`

`color(blue)(a_n=8n-22)`

God bless....I hope the explanation is useful.