# Thread: Solving an arithmetic sequence

1. ## Solving an arithmetic sequence

If we have a sequence with 20 terms, and the terms with odd numbers add up to 220 and the even ones add up to 250, what are the middle terms?

2. Originally Posted by Denny Crane
If we have a sequence with 20 terms, and the terms with odd numbers add up to 220 and the even ones add up to 250, what are the middle terms?
Any particular type of sequence?

CB

3. Arithmetic, by the title of his post...

4. Originally Posted by Prove It
Arithmetic, by the title of his post...
One can't rely on the subject line to indicate anything about a question, but lets assume it does.

Assuming we number the first term 1:

We have the sequence with initial term a and common difference d, 20 terms of which sum to 470, and a sequence with initial term a and common difference 2d, 10 terms of which sum to 220.

The summation formula applied to these two give two equations in two unknowns a and d, which can be solved to give a and d and then the middle terms of the 20 term series.

CB

5. Hello, Denny Crane!

If we have an arithmetic sequence with 20 terms,
and the terms with odd numbers add up to 220 and the even ones add up to 250,
what are the middle terms?
We have: .first term $a_1$, common difference $d$, and $n = 20$ terms.

Evidently, half the terms are odd and half are even.
. . and we see that $d$ is odd.

Since $\sum\text{odd} \:<\:\sum\text{even}$, the sequence begins with an odd number.

We have: . $\sum\text{odd} \:=\:a_1 + a_3 + a_5 + \hdots + a_{19} \;=\;220$

This is a sequence with: . $\begin{Bmatrix}\text{first term }a_1 \\ \text{common difference }2d \\ n = 10\text{ terms}\end{Bmatrix}$

Hence: . $\sum\text{odd} \:=\:\frac{10}{2}\bigg[2\left(a_1\right) + 9(2d)\bigg] \:=\:220 \quad\Rightarrow\quad a_1 + 9d \:=\:22$ .[1]

We have: . $\sum\text{even} \:=\:a_2 + a_4 + a_6 + \hdots + a_{20} \;=\;250$

This is a sequence with: . $\begin{Bmatrix}\text{first term } a_2 \:=\:a_1+d \\ \text{common difference }2d \\ n = 10\text{ terms} \end{Bmatrix}$

Hence: . $\sum\text{even} \:=\:\frac{10}{2}\bigg[2\left(a_1+d\right) + 9(2d)\bigg] \:=\:250 \quad\Rightarrow\quad a_1 + 10d \:=\:25$ .[2]

Subtract [1] from [2]: . $\boxed{d \:=\:3} \quad\Rightarrow\quad\boxed{a \:=\:-5}$

The sequence is: . $\text{-}5, \,\text{-}2, \:1, \:4, \:7 \;\hdots\; 52$

6. Originally Posted by Soroban
Hello, Denny Crane!

We have: .first term $a_1$, common difference $d$, and $n = 20$ terms.

Evidently, half the terms are odd and half are even.
. . and we see that $d$ is odd.

Since $\sum\text{odd} \:<\:\sum\text{even}$, the sequence begins with an odd number.

Snip...

You do the student a disservice by depriving them of the opportunity to do at least part of their problem themselves.

I know a complete solution may well be what the student wants, but is it what they need?

CB

7. Yeah. that's my specialty . . . doing damage to young minds.

Sorry, everyone!