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 $r={\displaystyle \frac{a_2}{a_1}}={\displaystyle \frac{0.1}{1}}=0.1={\displaystyle \frac{1}{10}}$, is

$$a_n={\displaystyle \frac{1}{{10}^{n-1}}}$$

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

$$P_m=\prod _{n=1}^m{\displaystyle \frac{1}{{10}^{n-1}}}=1^{10}\cdot {10}^{-45}$$

but we know that:

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

And by comparing both expressions, we have:

$${10}^{-45}=\sqrt{({\displaystyle \frac{1}{{10}^{m-1}}}{)}^m}$$

And by isolating the variable of this rational equation:

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

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

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

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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