Problems from Product of n terms of a geometric progression

Development:

The general term of the succession with the first term $$a_1=1$$ and ratio $$r=\dfrac{a_2}{a_1}=\dfrac{0.1}{1}=0.1=\dfrac{1}{10}$$, is

$$$a_n=\dfrac{1}{10^{n-1}}$$$

We want to find a natural $$m$$ in such a way that the product of the $$m$$ first terms of the succession is $$10^{-45}$$, that is to say, that

$$$P_m=\prod_{n=1}^m \dfrac{1}{10^{n-1}} = 1^{10}\cdot 10^{-45}$$$

but we know that:

$$$P_m=\sqrt{(a_1\cdot a_m)^m}=\sqrt{ \Big(1\cdot \dfrac{1}{10^{m-1}}\Big)^m}$$$

And by comparing both expressions, we have:

$$$10^{-45}= \sqrt{ \Big(\dfrac{1}{10^{m-1}}\Big)^m}$$$

And by isolating the variable of this rational equation:

$$$\Big(\dfrac{1}{10^{m-1}}\Big)^{\frac{m}{2}}=10^{-45} \Rightarrow \dfrac{1}{10^{\frac{m}{2}(m-1)}}=\dfrac{1}{10^{45}} \Rightarrow$$$

$$$10^{\frac{m(m-1)}{2}}=10^{45} \Rightarrow \dfrac{m^2-m}{2}=45 \Rightarrow$$$

$$$m^2-m-90=0$$$

So we only have to solve this equation of the second grade:

$$$m^2-m-90=0 \Rightarrow m=\{10,-9\}$$$

We know that $$m$$ must be a positive integer, so the solution is $$m=10$$.

Solution:

It is necessary to add the $$10$$ first terms.

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Development:

The general term of this progression is $$a_n=\sqrt{7}\cdot r^{n-1}$$ since we don't know the value of the ratio, but we do know that the first term is $$\sqrt{7}$$.

On the other hand, the product of the first six terms is $$\sqrt{7^{21}}$$, and if we apply the formula, we have:

$$$P_6=\sqrt{(a_1\cdot a_6)^6}=\sqrt{(\sqrt{7}\cdot\sqrt{7}\cdot r^5)^6}= \sqrt{7^6\cdot r^{30}}= 7^3\cdot r^{15}$$$

So,

$$$7^3\cdot r^{15} = \sqrt{7^{21}} \Rightarrow r^{15}=\sqrt{7^{15}} \Rightarrow r^{15}=(\sqrt{7})^{15} \Rightarrow r=\sqrt{7}$$$

Therefore, the general term of the succession is: $$$a_n=\sqrt{7^n}$$$ So, $$$a_1=\sqrt{7}, a_2=7, a_3=7\sqrt{7}, a_4=49, a_5=49\sqrt{7}$$$

Solution:

$$a_1=\sqrt{7}, a_2=7, a_3=7\sqrt{7}, a_4=49, a_5=49\sqrt{7}$$

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