Find the arithmetic progression where $a_7 \cdot a_8 = 1326$ and $a_1 \cdot a_{14} = 276$

Question

Find the arithmetic progression where $a_7 \cdot a_8 = 1326$ and $a_1 \cdot a_{14} = 276$

Right now I've only dealt with these problems when I have a sum of two members of the AP so I can use the formula for the sum of the AP. Right now, I don't know how to proceed. Could anyone help?

Answer 1

Ross Millikan gave you the correct general approach in the comments. Given that you're working with rather small numbers, another approach, specific to this question, is to factor $276$ and then look for two factors that differ by a multiple of $13$. That's because $13$ has to divide $a_{14}-a_1=(a_1+13d)-a_1$.

Answer 2

We have $a_n=a_1+(n-1)d$ for all $n\geq 1$. Thus the two pieces of information turn into the equations $a_1(a_1+13d)=276$ and $(a_1+6d)(a_1+7d)=1326$.

By subtracting the first equation from the second, we get $42d^2=1050\Longrightarrow d=\pm 5$. If $d=-5$, the first and second equations both give $a_1=-4,69$. If $d=5$, the first and second equations both give $a_1=4,-69$. So there are four solutions: $(a_1,d)=(-4,-5),(69,-5),(4,5),(-69,5)$.

Note that $(-4,-5)\equiv (-69,5)$ in the sense that they both refer to the sequence $\ldots ,-4,1,5,\ldots $, and $(69,-5)\equiv (4,5)$ in the sense that they both refer to the sequence $\ldots ,-1,4,9,\ldots $.