In the first question, we see that the common ratio is 5 and it starts at 7. That means every time you increment up, you multiply by seven, so:
[tex] 7(5)^{n-1} [/tex]
We have the n-1 in the exponent because we have to start out with 7/5 as our intial value because at n=1 we see we multiply something by 5 to get 7. Having [tex] 7(5)^{n-1} \text{is the same asv} \frac{7}{5}*5^n [/tex].
On 12, we can apply some of that logic here too. We know that the exponent is going to be "offset" by 1 so to speak, so we'll have:
[tex] t_6=s*r^5\\
t_10=s*r^9\implies \\
\frac{t_10}{t_6}=\frac{s*r^9}{s*r^6}=r^4 \implies \\
\sqrt[4]{\frac{t_10}{t_6}}=r [/tex]
[tex] \sqrt[4]{\frac{324}{4}}=\sqrt[4]{81}=3=r [/tex]
So our common ratio is 3. To find t_1, we can just do:
[tex] t_1=\frac{t_6}{r^5}=\frac{4}{3^5}=\frac{4}[243} [/tex]
