Respuesta :
hmmm [tex]\bf -3~,~-\cfrac{3}{2}~,~-\cfrac{3}{4}~,~-\cfrac{3}{8}~,~-\cfrac{3}{16}[/tex]
now, if you notice the terms there.... since we know is a geometric sequence/serie, we can tell there's a "common multiplier" namely a "common ratio", and if you divide any two terms, the latter by the former, the quotient you get is, yes, you guessed it, is the "common ratio", let's check about.
[tex]\bf -\cfrac{3}{2}\div -3\implies \cfrac{1}{2}\qquad \qquad -\cfrac{3}{4}\div -\cfrac{3}{2}\implies \cfrac{1}{2}[/tex]
so, as you can see, the "common ratio" is then, just 1/2.
now, for a geometric serie, whose common ratio is a fraction, namely less than 1 and greater than 0, then
[tex]\bf \textit{sum of an infinite geometric serie}\\\\ S=\sum\limits_{i=0}^{\infty}~a_1\cdot r^i\implies \cfrac{a_1}{1-r}\quad \begin{cases} a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ \qquad 0\ \textless \ |r|\ \textless \ 1\\ ----------\\ a_1=-3\\ r=\frac{1}{2} \end{cases} \\\\\\ S=\cfrac{-3}{1-\frac{1}{2}}\implies S=\cfrac{-3}{\frac{1}{2}}\implies S=-6[/tex]
now, if you notice the terms there.... since we know is a geometric sequence/serie, we can tell there's a "common multiplier" namely a "common ratio", and if you divide any two terms, the latter by the former, the quotient you get is, yes, you guessed it, is the "common ratio", let's check about.
[tex]\bf -\cfrac{3}{2}\div -3\implies \cfrac{1}{2}\qquad \qquad -\cfrac{3}{4}\div -\cfrac{3}{2}\implies \cfrac{1}{2}[/tex]
so, as you can see, the "common ratio" is then, just 1/2.
now, for a geometric serie, whose common ratio is a fraction, namely less than 1 and greater than 0, then
[tex]\bf \textit{sum of an infinite geometric serie}\\\\ S=\sum\limits_{i=0}^{\infty}~a_1\cdot r^i\implies \cfrac{a_1}{1-r}\quad \begin{cases} a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ \qquad 0\ \textless \ |r|\ \textless \ 1\\ ----------\\ a_1=-3\\ r=\frac{1}{2} \end{cases} \\\\\\ S=\cfrac{-3}{1-\frac{1}{2}}\implies S=\cfrac{-3}{\frac{1}{2}}\implies S=-6[/tex]
Answer: -6.
Step-by-step explanation: Given geometric series is-
[tex]-3,~-\dfrac{3}{2}, ~-\dfrac{3}{4}, ~-\dfrac{3}{8},~-\dfrac{3}{16},...[/tex]
Here, the first term of the series, [tex]a=-3[/tex]
and the common ratio, [tex]r=\dfrac{1}{2}.[/tex]
Since it is an infinite series, so sum is given by
[tex]S=\dfrac{a}{1-r}\\\\\\\Rightarrow S=\dfrac{-3}{1-\frac{1}{2}}\\\\\\\Rightarrow S=\dfrac{-6}{2-1}\\\\\\\Rightarrow S=-6.[/tex]
Thus, the sum of the series is -6.
