Question

The second and ninth terms of an arithmetic sequence are 2 and 30, respectively. What is the fiftieth term?

Answer 1

`a_(50)=194`

Explanation:

Ok let's say:

`a_n= a_s+d(n-s)`

This is a characteristic of an arithmetic sequence, each term is separated by a common difference:

In this case:
`a_9= a_2+d(9-2)`

`30= 2+7d`

`28=7d`

`d=4`

So, the 5th term:
`a_(50)= a_2+d(50-2)`

`a_5= 2+4(48)`

`a_5=194`

Answer 2

`color(blue)(194)`

Explanation:

The nth term of an arithmetic sequence is given by:

`a+(n-1)d`

Where:

`bba` is the first term, `bbd` is the common difference and `bbn` is the nth term.

We have:

`a+(2-1)d=2 \ \ \ \[1]`

and

`a+(9-1)d=30 \ \ \ \[2]`

We need to find `bba` and `bbd`.

Solving `[1]` and `[2]` simultaneously:

Subtract `[1]` from `[2]`

`a-a+8d-d=30-2`

`7d=28=>d=4`

Substituting in `[1]`

`a+4=2=>a=-2`

So our general term is:

`-2+(n-1)4`

50th term will therefore be:

`-2+(50-1)4=194`

Answer 3

`a_(50)=194`

Explanation:

`"the n th term of an arithmetic sequence is"`

`•color(white)(x)a_n=a+(n-1)d`

`" where a is the first term and d the common difference"`

`a_2=a+d=2to(1)`

`a_9=a+8d=30 to(2)`

`(2)-(1)" gives"`

`7d=28rArrd=4`

`"substitute in "(1)a+4=2rArra=-4`

`rArra_(50)=-2+(49xx4)=-2+196=194`