1. ## Geometric progression

in a geometric progression a4 * a5=1/64
a1*a2*a3*a4....*a13=?

2. ## Re: Geometric progression

and these question's answer is 40?
In a Geometric progression :
a5 is 24 more than a3. => a5-a3=24
if terms are Integer and positive. a3+a5=?

3. ## Re: Geometric progression

Originally Posted by parmis
in a geometric progression a4 * a5=1/64
a1*a2*a3*a4....*a13=?
A geometric progression is one of the form $a, ar, ar^2, ar^3,\ldots$. So, suppose $a_0 = a, a_1 = ar, a_2 = ar^2, a_3 = ar^3, a_4 = ar^4, a_5 = ar^5$. Then $a_4 a_5 = (ar^4)(ar^5) = a^2r^9 = \dfrac{1}{64}$. So, $\prod_{n=1}^{13} a_n = a^{13}r^{91} = (a^2r^9)^6(ar^{37}) = \left(\dfrac{1}{64}\right)^6(ar^{37})$. You do not have enough information to solve this.

Edit: We can actually do a little better. Since $a\neq 0$, we can solve for $r$. $a^2r^9 = \dfrac{1}{64} \Rightarrow r^9 = (8a)^{-2} \Rightarrow r = (8a)^{-2/9}$. Then $\prod_{n=1}^{13} a_n = a^{13}r^{91} = a^{13}(8a)^{-182/9} = a^{-65/9}2^{-182/3}$. That's the closest I can come to a solution.

4. ## Re: Geometric progression

Originally Posted by parmis
and these question's answer is 40?
In a Geometric progression :
a5 is 24 more than a3. => a5-a3=24
if terms are Integer and positive. a3+a5=?

Terms are integers. So, $a_3 = ar^3, a_5 = ar^5$. Suppose $a=1$. Then $a_5-a_3 = r^5-r^3=24$ has $r=2$ as a solution. For any integer $a>1$, there are no integer solutions to $a(r^5-r^3) = 24$. So, that must be the correct solution. Then $a_3+a_5 = r^3+r^5 = 2^3+2^5 = 8 + 32 = 40$.

5. ## Re: Geometric progression

Hello, parmis!

There are at least 4 possible solutions . . .

$\text{In a geometric progression: }\;a_4\cdot a_5\,=\,\tfrac{1}{64}$

$\text{Find: }\:P \;=\;a_1\cdot a_2\cdot a_3 \cdots a_{13}$

We are given: . $a_4\cdot a_5 \,=\,\tfrac{1}{64}$

I found four cases: . $(a_4,a_5) \:=\:\left(\tfrac{1}{16},\tfrac{1}{4}\right),\; \left(\tfrac{1}{4},\tfrac{1}{16}\right),\;\left( \text{-}\tfrac{1}{16},\text{-}\tfrac{1}{4}\right),\;\left(\text{-}\tfrac{1}{4},\text{-}\tfrac{1}{64}\right)$

Take the first case: . $\begin{Bmatrix}a_4 &=& \frac{1}{16} \\ a_5 &=& \frac{1}{4} \end{Bmatrix}$

We find that: . $a = \tfrac{1}{4^5},\;r = 4$

Then: . $P \;=\;a_1\cdot a_2\cdot a_3\cdot a_4 \cdots a_{13}$

. . . . . . . . $=\; (a)(ar)(ar^2)(ar^3) \cdots (ar^{12})$

. . . . . . . . $=\; a^{13}r^{78}$

. . . . . $P \;=\;\left(\tfrac{1}{4^5}\right)^{13}(4)^{78} \;=\;\tfrac{1}{4^{65}}\cdot4^{78}$

. . . . . . . $=\;4^{13} \;=\;67,\!108,\!864$

You can work out the other three solutions.