Question

I need a little help on #13

Answer 1

`\bf \begin{array}{llll} term&value\\ \textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\ t_1\\ t_2\\ t_3&18\\ t_4&18\cdot r\to 18r\\ t_5&(18r)r\to 18r^2\\ t_6&(18r^2)r\to 18r^3\\ t_7&(18r^3)r\to 18r^4=1458\\ \end{array}\qquad thus \\\\\\ 18r^4=1458\implies r^4=\cfrac{1458}{18}\implies r=\sqrt[4]{\cfrac{1458}{18}} \\\\\\ now\implies t_n=t_1\cdot r^(n-1)`


`\bf \begin{cases} t_1=\textit{first term}\\ r=\textit{common ratio}\\ t_n=n^(th)\ term \end{cases} \\\\\\ \textit{we know the 7th term is 1458, thus} \\\\\\ t_7=t_1\cdot r^(7-1)\implies \cfrac{1458}{r^(7-1)}=t_1`

and just plug in there the "common ratio" or "r" you found, to get the first term

Answer 2

I believe that first question would be the answer