Find the first term and the common ratio of an infinite geometric series whose sum is $5$ and such that each term is $4$ times the sum of all the terms that follow it.
I used $a_{1}r^{3}=\frac{4[a_{1}(r^{3}-1)]}{r-1}$ infinite geometric series. Solving that I got the value of $r= -0.83$. Substituting to the formula of infinite GS, I have my $a_1= 9.15$. many thanks in advance.
We have $$\frac{a_1}{1-r}=5$$ and $$a_n=\frac{4a_{n+1}}{1-r}.$$ The second gives $$1=\frac{4r}{1-r}$$ or $$r=\frac{1}{5},$$ which gives $$\frac{a_1}{1-\frac{1}{5}}=5$$ and $$a_1=4$$