Question

I need help with question 2.

Answer 1

So.. hmmm if we look at the ratio, we can say that $\bf \cfrac{u_9}{u_4}=\cfrac{3^9}{3^4}\implies \cfrac{u_9}{u_4}=3^9\cdot 3^{-4} \implies \cfrac{u_9}{u_4}=3^5$  thus, our common ratio is really 3

so hmm what's the first term?  well, we know the fourth term is 135 or $\bf u_4=135$ , now, in a geometric sequence, we get the next term, by multiplying the current by the common ratio

what if, we divide the "next" term by the common ratio? we get the previous or "current" term

so $\bf \begin{array}{cllll} term&value\\ \textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\ u_4&135\\\\ u_3&\frac{135}{3}\to 45\\\\ u_2&\frac{45}{3}\to 15\\\\ u_1&\frac{15}{3}\to 5\\ \end{array}$

so the first term is 5 then

now, what's the Sum of the first 10 terms?

well, we know the first term and common ratio, let's use those fellows

$\bf S_n=u_1\left( \cfrac{1-r^n}{1-r} \right)\qquad \begin{cases} u_1=\textit{first term}\\ n=n^{th}\ term\\ r=\textit{common ratio}\\ --------\\ u_1=5\\ n=10\\ r=3 \end{cases}\\\\\\ S_{10}=5\left( \cfrac{1-3^{10}}{1-3}\right) \\\\\\ S_{10}=5\left( \cfrac{1-59049}{-2}\right)\implies S_{10}=5\left( \cfrac{-59048}{-2}\right)\\\\\\ \begin{array}{llcll} S_{10}=&5(&29524)\\ &\uparrow &\uparrow \\ &a&b \end{array}$