# Thread: Sequences and Series Help Needed

1. ## Sequences and Series Help Needed

Steven takes an arithmetic sequence with common difference d and a geometric sequence with common ratio r and adds them together, term by term, to form a new sequence (i.e., the first term of a new sequence is the sum of the first terms of the arithmetic and geometric sequences, etc). If the first three terms of the new sequence are, in order, 3, 8, and 15, and d and r are both known to be positive integers, what is the sum of the possible values of d?

2. ## Re: Sequences and Series Help Needed

Hello, Espionage!

This is a very "busy" problem . . .

Steven takes an arithmetic sequence with common difference $\displaystyle d$
and a geometric sequence with common ratio $\displaystyle r$,
and adds them together, term by term, to form a new sequence.

If the first three terms of the new sequence are, in order, 3, 8, and 15,
and $\displaystyle d$ and $\displaystyle r$ are both positive integers,
what is the sum of the possible values of $\displaystyle d$?

$\displaystyle \begin{array}{|c||c|c|c|} \hline \text{A.S.} & a & a+d & a+2d \\ \text{G.S.} & a & ar & ar^2 \\ \hline \hline \text{Sum} & 2a & ar+a+d & ar^2 + a + 2d \\ \hline\end{array}$

We are told: .$\displaystyle \begin{Bmatrix}2a \:=\:3 & [1] \\ ar +a+d \:=\:8 & [2] \\ ar^2+a+2d \:=\:15 & [3] \end{Bmatrix}$

From [1]: .$\displaystyle 2a \,=\,3 \quad\Rightarrow\quad a \,=\,\tfrac{3}{2}$

Substitute into [2]: .$\displaystyle \tfrac{3}{2}r + \tfrac{3}{2} + d \:=\:8 \quad\Rightarrow\quad 3r + 2d \:=\:13 \;\;[4]$

Substitute into [3]: .$\displaystyle \tffrac{3}{2}r^2 + \tfrac{3}{2} + 2d \:=\:15 \quad\Rightarrow\quad 3r^2 + 4d \:=\:27\;\;[5]$

$\displaystyle \begin{array}{cccccc}\text{We have [5]:} & 3r^2 + 4d &=& 27 \\ 2\times [4]\!: & 6r + 4d &=& 26 \end{array}$

$\displaystyle \text{Subtract: }\quad 3r^2 - 6r \:=\:1 \quad\Rightarrow\quad 3r^2 - 6r -1 \:=\:0$

Quadratic Formula: .$\displaystyle r \:=\:\frac{6\pm\sqrt{36+12}}{6} \:=\:\frac{3\pm2\sqrt{3}}{3}$

Substitute into [4]: .$\displaystyle 3\left(\frac{3\pm2\sqrt{3}}{3}\right) + 2d \:=\:13 \quad\Rightarrow\quad d \:=\:5 \pm\sqrt{3}$

The sum of the possible $\displaystyle d$-values is: .$\displaystyle \left(5 + \sqrt{3}\right) + \left(5-\sqrt{3}\right) \;=\;10$