Find the first term in the arithmetic sequence if the 9th term is 55 and the 3rd term is 19

Answer 1

We know that the n^th term for an arithmetic sequence is

a_n = a + (n − 1)d

Where a is the first term of the sequence and d is the common difference

Given: a₃ = a + (3 − 1)d = 19

= a + 2d = 19

and a₉ = a + (9 − 1)d = 55

= a + 8d = 55

Now solve for d for both equations

d = (19 − a) / 2 and d = (55 − a) / 8

Then we set them equal to each other and solve for a

(19 − a) / 2 = (55 − a) / 8

which can be written as

(19 / 2) − (a / 2) = (55 / 8) − (a / 8) *Add (a / 8) to both sides

*Subtract (19 / 2) from both sides

−(3 / 8)a = −(21 / 8) *Now multiply −(8 / 3) to both sides

And finally you will get that

a = 7

To check our work, we plug in 7 for a for either equation and any d as well

For this case we will use a₃ and d = (19 − a) / 2 = (19 − 7) / 2 = 6

So a₃ = a + 2d = 7 + 2(6) = 7 + 12 = 19

Hence Proved