Question

Determinati al 9 lea termen al unei progresii geometrice stiind ca ratia este egala cu 1/3 si primul numar este 243

Answer 1

b1=243
q=1/3
b9=?
$b_{9}=b_{1}*q^{8}\ \textless \ =\ \textgreater \ b_{9}=243* (\frac{1}{3}) ^{8}=\ \textgreater \ b_{9}=243* \frac{1}{6561} =\ \textgreater \ b_{9}= \frac{1}{27}$

Answer 2

$\displaystyle b_1=243,~q= \frac{1}{3} ,~b_9=? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{b_n=b_1 \cdot q^{n-1}} \\ \\ b_9=243 \cdot \left( \frac{1}{3} \right)^{9-1} \Rightarrow b_9=243 \cdot \left( \frac{1}{3} \right)^8 \Rightarrow b_9=243 \cdot \frac{1^8}{3^8} \Rightarrow \\ \\ \Rightarrow b_9=243 \cdot \frac{1}{6561} \Rightarrow b_9= \frac{243}{6561} ^{(243} \Rightarrow \boxed{b_9= \frac{1}{27} }$