Question

You have to find the common difference, the term named in the problem, and the explicit formula.

Answer 1

Question 1

Given the sequence

31, 61, 91, 121,...

An arithmetic sequence has a constant difference 'd' and is defined by

`a_n=a_1+\left(n-1\right)d`

computing the differences of all the adjacent terms

61 - 31 = 30, 91 - 61 = 30, 121 - 91 = 30

The difference between all the adjacent terms is the same and equal to

d = 30

As the first term of the sequence is:

a₁ = 31

now substituting a₁ = 31 and d = 30 in the nth term of the sequence

`a_n=a_1+\left(n-1\right)d`

`a_n=30\left(n-1\right)+31`

`a_n=30n+1`

Now, putting n = 36 to determine the 36th term

`a_n=30n+1`

`a_(36)=30\left(36\right)+1`

`a_(36)=1080+1`

`a_(36)=1081`

Thus, the 36th term is:

`a_(36)=1081`

Question 2

Given the sequence

-34, -44, -54, -64, ...

An arithmetic sequence has a constant difference 'd' and is defined by

`a_n=a_1+\left(n-1\right)d`

computing the differences of all the adjacent terms

`-44-\left(-34\right)=-10,\:\quad \:-54-\left(-44\right)=-10,\:\quad \:-64-\left(-54\right)=-10`

The difference between all the adjacent terms is the same and equal to

`d=-10`

As the first term of the sequence is:

a₁ = -34

now substituting a₁ = -34 and d = -10 in the nth term of the sequence

`a_n=a_1+\left(n-1\right)d`

`a_n=-10\left(n-1\right)-34`

`a_n=-10n-24`

Now, putting n = 26 to determine the 36th term

`a_n=-10n-24`

`a_(26)=-10\left(26\right)-24`

`a_(26)=-260-24`

`a_(26)=-284`

Thus, the 26th term is:

`a_(26)=-284`