Question

The third term of an arithmetic sequence is 14, and the ninth term is -1. How do you find the first four terms of the sequence?

Answer 1

`19" "16.5" "14" "11.5" " ....`

Explanation:

You can write an equation (or formula) for each term in an AP.

`T_n = a + d(n-1)`

"The third term is 14" can be written with the formula as:

`color(blue)(T_3 = a +2d = 14)" "rarr(n-1) = 3-1 = 2`

"The ninth term is -1 " can be written as:

`color(blue)(T_9= a +8d= -1)" "rarr(n-1) = 9-1 =8`

Now there are two equations with two unknowns.
Solve them simultaneously (as a system).

`color(blue)(T_9= a +8d= -1`............................A
`color(blue)(T_3 = a +2d = +14)` ........................B

A-B: `" "6d = -15`
`" "d = -15/6 = -5/2" "larr`this is the value for d

`color(blue)(T_3 = a +2((-5)/2) = +14)`

`a -5 = 14`
`a = 19" "larr` this is the value for a

This sequence is given by `color(red)(T_n = 19 -5/2d)`

`T_1 = a = 19`
For `T_2: n = 2 rarr 19-5/2(1) = 16.5`
`T_3: 16.5-5/2 = 14`
`T_4: 14-5/2 = 11.5`

Each term is 2.5 less than the previous term:

`19" "16.5" "14" "11.5" "9 ....`