Question

calcule a soma dos cincos primeiros termos da PG (45, 15, 5)

Answer 1

$Sabemos\ que:\\\\ PG=(45,15,5,...) \ \ \| \ \ n=5\\ q= \frac{a_{n}}{a_{n-1}} \ \ \| \ \ S_n= \frac{a_1.(q^n-1)}{q-1} \\\\ Logo:\\\\ q= \frac{a_2}{a_1}=\frac{15}{45}=\frac{1}{3}\\\\ S_n= \frac{45.({[\frac{1}{3}]}^5-1)}{\frac{1}{3}-1}\\\\ S_n= \frac{45.(\frac{1}{243}-1)}{\frac{1-3}{3}}\\\\ S_n=\frac{45.(\frac{1-243}{243})}{\frac{-2}{3}}\\\\ S_n=45.(\frac{-242}{243}).\frac{3}{-2}\\\\ S_n=45.(\frac{121}{81})=5.(\frac{121}{9})\\\\ S_n= \frac{605}{9} \approx \| \ 67,22 \ \|$