Question

If the 100th term of an arithmetic sequence is 609, and its common difference is 6, then
its first term a1=
It’s second term a2=
It’s third term a3=

Answer 1

Answer:

15, 21, 27

Step-by-step explanation:

$a_n}$ = $a_{1}$ + (n-1)d  

$a_{n}$ is the number in the sequence we are looking for

$a_{1}$  This is the first term in the sequence that we do not know and we are looking for

n stands for the number of the term and d is the common difference.  We will put in all that we know and solve for the first term.

609 = $a_{1}$ + (100-1)6  

609 =$a_{1}$ + (99)6

609 = $a_{1}$ + 594  Subtract 594 from both sides of the equation

15 = $a_{1}$

Now that we know that the first term is 15 we just add 6 to get the next term which is 21 and then add 6 again to get the last term 27.