Question

The fifth term of an arithmetic sequence is 11 and the tenth term is 41. What is the first term?

Answer 1

In this sequence, the first term is
`a_1` and every successive term is determined by

`a_n=a_(n-1)+d`

where
`d` is the common difference between terms. We have

`a_(11)=a_(10)+d=a_(9)+2d=\cdots=a_5+6d`

so that

`41=11+6d\implies6d=30\implies d=5`

Then

`a_5=a_4+5=a_3+2\cdot5=\cdots=a_1+4\cdot5`

`\implies11=a_1+20\implies a_1=-9`

Answer 2

The first term is -13.

Explanation:

The general rule of an arithmetic sequence is the following:

`a_(n+1) = a_(n) + d`

In which d is the common diference between each term.

This is the case going from one term to the next. However, when, as in this problem, we have the fifth and the tenth term, this formula can be expanded, as the following way:

`a_(n + m) = a_(n) + m*d`

So

`a_(10) = a_(5) + 5*d`

`41 = 11 + 5d`

`5d = 30`

`d = 6`

The common diference is 6.

To find the first term, we do:

`a_(5) = a_(1) + 4*d`

`11 = a_(1) + 4*6`

`a_(1) = -13`

The first term is -13.