# Thread: The sum of the first three terms of an arithmetic sequence is 12. The sum of their s

1. ## The sum of the first three terms of an arithmetic sequence is 12. The sum of their s

Question -
The sum of the first three terms of an arithmetic sequence is 12. The sum of their squares is 66. Find the fourth term

I know just by trial an error that the first three terms are 1, 4 and 7. therefore the 4th term is 10.
Although I am able to figure this out, there IS an algebraic way of determining the values to prove that these ar , in fact, the numbers of the sequence.

I have started along the lines of
a + (a+d)^2 + (a+2d)^2 = 66
a + a +(a+d) + (a+d) + (a+2d) + (a+2d) = 66
6a+6d = 66

but im not sure where to go from here. i know im on the right track i just cant figure this out. please help !?

2. Originally Posted by Iainwb.13
Question -
The sum of the first three terms of an arithmetic sequence is 12. The sum of their squares is 66. Find the fourth term

I know just by trial an error that the first three terms are 1, 4 and 7. therefore the 4th term is 10.
Although I am able to figure this out, there IS an algebraic way of determining the values to prove that these ar , in fact, the numbers of the sequence.

I have started along the lines of
a + (a+d)^2 + (a+2d)^2 = 66
a + a +(a+d) + (a+d) + (a+2d) + (a+2d) = 66
6a+6d = 66

but im not sure where to go from here. i know im on the right track i just cant figure this out. please help !?
The terms will be: $a, a+d, a+2d, a+3d$

$a + (a+d) + (a+2d) = 3a + 3d = 12 \qquad \rightarrow \qquad a + d = 4$

$a^2 + (a+d)^2 + (a+2d)^2 = a^2 + a^2 + 2ad + d^2 + a^2 + 4ad + 4d^2 = 3a^2 + 6ad + 5d^2 = 66$

So now you have two equations and two unknowns. Plug in the first equation $a+d = 4$ into the second equation. Then you can solve for $a$ and $d$

Then your fourth term is just $a+3d$

3. Hello, Iainwb.13!

You had a good start . . .

The sum of the first three terms of an arithmetic sequence is 12.
The sum of their squares is 66.
Find the fourth term.. .
. . . There are two answers.

The first three terms are: . $a,\:a+d,\:a+2d$

Their sum is 12:
. . $a + (a+d) + (a+2d) \:=\:12 \quad\Rightarrow\quad 3a + 3d \:=\:12 \quad\Rightarrow\quad d \:=\:4-a$ .[1]

The sum of their squares is 66:
. . $a^2 + (a+d)^2 + (a+2d)^2 \:=\:66 \quad\Rightarrow\quad 3a^2 + 6ad + 5d^2 \:=\:66$ .[2]

Substitute [1] into [2]: . $3a^2 + 6a(4-a) + 5(4-a)^2 \:=\:66$

. . which simplifies to: . $a^2 - 8a + 7 \:=\:0$

. . which factors: . $(a-1)(a-7) \:=\:0$

. . and has roots: . $a \;=\;1,\:7$

Substitute into [1]: . $d \;=\;3,\:\text{-}3$

There are two possible sequences . . .

. . $\begin{array}{ccccccc}a = 1,\;d = 3\!: & 1,\:4,\:7 & \text{ The fourth term is 10.} \\

a = 7,\;d = \text{-}3\!: & 7,\:4,\:1 & \text{ The fourth term is -2.}\end{array}$

4. Thank you soo much .
I tried doing it by myself before i checked your answers and i did end up with exactly that

thank you so much, your help was greatl appreciated