Question

Please help with these problems.

Answer 1

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:

`7(5)^(n-1)`

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
`7(5)^(n-1) \text{is the same asv} (7)/(5)*5^n`.

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:

`image`

`\sqrt[4]{(324)/(4)}=\sqrt[4]{81}=3=r`

So our common ratio is 3. To find t_1, we can just do:

`t_1=(t_6)/(r^5)=(4)/(3^5)=\frac{4}[243}`