# Arithmetic and Geometric Sequences Relationship

This is a question from our long quiz in math that we had just a while ago. And I wasn't able to answer it, if I could have had then I would have got a perfect score.

4 positive integers form an arithmetic progression. If we subtract 2, 6, 7, and 2, respectively, from the 4 numbers, it forms a geometric progression. What are the 4 numbers?

Firstly, the original numbers must be "larger" than $\{5,6,7,8\}$ since we will subtract 6 and 7.

Depending on what kind of class this quiz took place, there might be shortcuts where one can make some more "educated guesses".

Observe that the subtraction of $\{2,6,7,2\}$ are all small, so they cannot create large differences (and we have positive numbers to begin with). A reasonable guess is that the resultant geometric progression cannot possibly have a large common ratio $r$. One might try out some nice small whole numbers like $r = 2$.

You write down guesses of geometric progression for different initial terms like $\{1, 2, 4, 8\}$, $\{2, 4, 8, 16\}$, and $\{3, 6, 12, 24\}$, then reverse the operation and add $\{2,6,7,2\}$ to see if they are arithmetic.

Voila, you see that $\{3, 6, 12, 24\}$ works nicely, coming from an original arithmetic progression $\{5, 12, 19,26\}$.

In general, the following is how one solves the question regardless if the numbers are nice.

Denote the original 4 numbers in arithmetic progression as $\{a, a+d, a+2d, a+3d\}$, then the geometric progression (with same initial $a$ and ratio $r$) after the subtraction gives \begin{align*} r = \frac{ a + d - 6 }{ a -2 }\quad\text{as Eq.(1),} && r = \frac{ a + 2d - 7 }{ a + d - 6 } \quad\text{as Eq.(2),} && r = \frac{ a + 3d - 2 }{ a + 2d - 7} \quad\text{as Eq.(3)} \end{align*} It is the common $r$ for all 3 equations by definition. We have 3 unknowns $\{a, d, r\}$ and 3 equations. Let's solve them.

Cross multiply Eq.(1) and Eq.(2) \begin{align*} (a-2)(a+2d-7) &= (a + d -6)^2 \\ a^2 + 2ad -9a - 4d + 14 &= a^2 + d^2 +36 -12 a - 12 d + 2 a d \\ \color{magenta}{3a} &= \color{magenta}{d^2 - 8d + 22} \end{align*} Cross multiply Eq.(2) and Eq.(3) \begin{align*} (a+d-6)(a+3d-2) &= (a + 2d -7)^2 \\ a^2 + 3d^2 + 12 + 4ad -8a - 20d &= a^2 + 4d^2 + 49 -14 a - 28 d + 2 a d \\ \color{magenta}{6a} &= \color{magenta}{d^2 - 8d + 37} \end{align*} The two magenta equations give $3a = 15$ therefore $a = 5$.

Substituting back to either one of the magenta equations we get $d^2 - 8d +7 = 0 \implies (d-1)(d-7)=0$, hence $d = 1$ or $d = 7$

If $d = 1$, then the original arithmetic sequence is $5,6,7,8$, which after the subtraction is $\{3,0,0,6\}$. This is clearly not a solution. The $d = 1$ is extraneous where $a+d-6=0$ and $a+2d-7 = 0$ invalidate the cross multiplication.

For $d = 7$, we have the original arithmetic progression as $\{5, 12, 19,26\}$, which upon the subtraction is the geometric progression $\{3, 6, 12, 24\}$. This is our solution.

• Thank you so much, Mr. David. Great help. Apr 30, 2018 at 8:08

GP $\lbrace a, ar, ar^2, ar^3\rbrace$.

Adding $\lbrace 2,6,7,2\rbrace$ to the GP gives an AP (AP1).

Subtracting $2$ from each term of AP1 gives another AP (AP2), with the same common difference $d$.

Altenratively, AP2 can be derived by adding $\lbrace 0,4,5,0\rbrace$ to the GP.

$$\begin{array} \hline \text{GP}: &&a \; \; \; &ar &ar^2 &ar^3\\ \text{add}: &&0 &4 &5 &0\\ \hline \text{AP2}: &&a &ar+4 & ar^2+5 &ar^3\\ &=&a &a+d &a+2d &a+3d\\ \hline \end{array}$$

For AP2,

\begin{align} 2(ar+4)&=a+(ar^2+5)\\ a(r-1)^2&=3\tag{1} \end{align} and \begin{align} 2(ar^2+5)&=(ar+4)+ar^3\\ ar(r-1)^2&=6\tag{2} \end{align}

which gives $r=2, a=3, d=7$.

Hence

• GP is $\lbrace 3,6,12,24\rbrace$, and

• AP (AP1) is $\color{red}{\lbrace 5,12,19, 26\rbrace}$.

Note that AP2 is $\lbrace 3,10,17, 24\rbrace$.

• Good stuff. I think this is the intended solution when the problem was designed. May 31, 2018 at 5:12
• @LeeDavidChungLin Thanks :) May 31, 2018 at 9:01