1. ## Sequence

In the sequence $\displaystyle a_1,a_2,\ldots, a_{80}$:
• $\displaystyle a_i>0\,\,\forall 1\le i\le 80$
• $\displaystyle a_i=a_{i-1}a_{i+1}\,\,\forall 2\le i\le 79$
• $\displaystyle \prod_{i=1}^{40}a_i=8=\prod_{i=1}^{80}a_i$

What can $\displaystyle a_1,a_2,\ldots , a_8$ be?

2. Hello james_bond
Originally Posted by james_bond
In the sequence $\displaystyle a_1,a_2,\ldots, a_{80}$:
• $\displaystyle a_i>0\,\,\forall 1\le i\le 80$
• $\displaystyle a_i=a_{i-1}a_{i+1}\,\,\forall 2\le i\le 79$
• $\displaystyle \prod_{i=1}^{40}a_i=8=\prod_{i=1}^{80}a_i$

What can $\displaystyle a_1,a_2,\ldots , a_8$ be?
Re-write the recurrence relation as $\displaystyle a_{i+1}=\frac{a_i}{a_{i-1}}$, and then look at the first few terms, given $\displaystyle a_1$ and $\displaystyle a_2$:

$\displaystyle a_3=\frac{a_2}{a_1}$

$\displaystyle a_4=\frac{a_3}{a_2}=\frac{1}{a_1}$

$\displaystyle a_5=\frac{a_4}{a_3}=\frac{1}{a_2}$

$\displaystyle a_6=\frac{a_5}{a_4}=\frac{a_1}{a_2}$

$\displaystyle a_7 = ... = a_1$

$\displaystyle a_8 = ... = a_2$

So the sequence repeats after six terms.

Then note that $\displaystyle \prod_{i=1}^6a_i=a_1a_2\cdot\frac{a_2}{a_1}\cdot\f rac{1}{a_1}\cdot\frac{1}{a_2}\cdot\frac{a_1}{a_2}= 1$

$\displaystyle \Rightarrow \prod_{i=1}^{36}a_i=1$

$\displaystyle \Rightarrow \prod_{i=1}^{40}a_i=a_1a_2a_3a_4=\frac{{a_2}^2}{a_ 1}=8$

$\displaystyle \Rightarrow a_1=\frac{{a_2}^2}{8}$ (1)

Similarly $\displaystyle \prod_{i=1}^{78}a_i=1$

$\displaystyle \Rightarrow \prod_{i=1}^{80}a_i=a_1a_2=8$

$\displaystyle \Rightarrow {a_2}^3=64$ from (1)

$\displaystyle \Rightarrow a_2=4$ and $\displaystyle a_1 = 2$

So the sequence is $\displaystyle 2, 4, 2, \frac{1}{2}, \frac{1}{4},\frac{1}{2}, 2, 4, ...$