Question

Find the 29th term of an arithmetic sequence with a1 = 2 and d =5

Answer 1

Formula: an=a1 + (n-1) x d

a1=2

d=5

n=29

plug in the values: an= 2 + (29-1) x 5 = 142

SO THE ANSWER IS : 142

Answer 2

Arithmetic Sequence is a sequence of numbers such that the difference between each number is constant. The formula for the arithmetic sequence is given by;

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

where An is the last term

A₁ is the first term

n is the number of terms (or number in the sequence)

d is the common difference

In our problem we are given the first term (a₁) = 2, a common difference of 5 (d = 5), and the number of terms which is 29 (n = 29).

Now in order to find the 29th term of the sequence (a₂₉), we just need to follow the formula;

`\begin{gathered} a_(29)=a_1+(n-1)d_{} \\ a_(29)=2_{}+(29-1)5 \\ a_(29)=2_{}+(28)5 \\ a_(29)=2_{}+140 \\ a_(29)=142 \end{gathered}`

Therefore the 29th term of the arithmetic sequence is 142.