Sequence in term

Question

what is the 125th term in the sequence 3,9.15.21

Answer 1

You could keep counting by sixes until you get to the 125th term of the sequence. Good luck with that.

Or you could use the fact that this is an arithmetic sequence with a first term of 3 and a common difference of 6. Then realize that you can determine the 125th term by starting with 3 and adding on 124 differences (124)(6).

The calculation looks like this:

125th term = 3 + (125-1)6 = 3 + 124(6) = 3 + 744 = 747.

The formula for the nth term of an arithmetic sequence::

a_n = a₁ + (n-1)d

For this particular problem:

a_n= 3 + (n-1)6

=3 + 6n -6

=6n-3

If you plug 125 in for n:

6(125) - 3 = 750 - 3 = 747

Take a look here to see ways to determine this on Desmos:

desmos.com/calculator/sgis2qoehd