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?
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
Hello, Denny Crane!
We have: .first term $\displaystyle a_1$, common difference $\displaystyle d$, and $\displaystyle n = 20$ terms.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?
Evidently, half the terms are odd and half are even.
. . and we see that $\displaystyle d$ is odd.
Since $\displaystyle \sum\text{odd} \:<\:\sum\text{even}$, the sequence begins with an odd number.
We have: .$\displaystyle \sum\text{odd} \:=\:a_1 + a_3 + a_5 + \hdots + a_{19} \;=\;220$
This is a sequence with: .$\displaystyle \begin{Bmatrix}\text{first term }a_1 \\ \text{common difference }2d \\ n = 10\text{ terms}\end{Bmatrix}$
Hence: .$\displaystyle \sum\text{odd} \:=\:\frac{10}{2}\bigg[2\left(a_1\right) + 9(2d)\bigg] \:=\:220 \quad\Rightarrow\quad a_1 + 9d \:=\:22$ .[1]
We have: .$\displaystyle \sum\text{even} \:=\:a_2 + a_4 + a_6 + \hdots + a_{20} \;=\;250$
This is a sequence with: .$\displaystyle \begin{Bmatrix}\text{first term } a_2 \:=\:a_1+d \\ \text{common difference }2d \\ n = 10\text{ terms} \end{Bmatrix}$
Hence: .$\displaystyle \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]: .$\displaystyle \boxed{d \:=\:3} \quad\Rightarrow\quad\boxed{a \:=\:-5}$
The sequence is: .$\displaystyle \text{-}5, \,\text{-}2, \:1, \:4, \:7 \;\hdots\; 52$