Answer:
a₄ = 8
Step-by-step explanation:
The sum to n terms of a geometric sequence is
[tex]S_{n}[/tex] = [tex]\frac{a(1-r^{n}) }{1-r}[/tex]
where a is the first term and r the common ratio, thus
[tex]S_{4}[/tex] = [tex]\frac{a(1-2/3)^{4} }{1-\frac{2}{3} }[/tex] = 65, that is
[tex]\frac{a(1-\frac{16}{81}) }{\frac{1}{3} }[/tex] = 65
3a × [tex]\frac{65}{81}[/tex] = 65
[tex]\frac{195a}{81}[/tex] = 65 ( multiply both sides by 81 )
195a = 5265 ( divide both sides by 195 )
a = 27
Hence
a₂ = 27 × [tex]\frac{2}{3}[/tex] = 18
a₃ = 18 × [tex]\frac{2}{3}[/tex] = 12
a₄ = 12 × [tex]\frac{2}{3}[/tex] = 8
A geometric series is the series in which the ratio of the two successive terms is constant for the whole series. The 4th term of the given geometric series is 8.
The sum of the four terms of a geometric series is 65.
The ratio of the geometric series is 2/3.
A geometric series is the series in which the ratio of the two successive terms is constant for the whole series.
The formula to fine the sum of the nth term of the geometric series can be given as,
[tex]S_n=\dfrac{a(1-r^n)}{1-r}[/tex]
Here [tex]a[/tex] is the first term of the series and [tex]r[/tex] is the common ratio.
Put the values in the above formula,
[tex]\begin{aligned}\\S_4&=\dfrac{a(1-(\dfrac{2}{3} )^4)}{1-\dfrac{2}{3} }\\65&=\dfrac{a(1-(\dfrac{2}{3} )^4)}{\dfrac{1}{3} }\\\\65\times \dfrac{1}{3} &={a\times0.80247}{\\\end[/tex]
Solve for the a we get,
[tex]a=27[/tex]
As the first term is 27 and common ratio is 2/3. Thus,
[tex]a_2=27\times\dfrac{2}{3} =18[/tex]
[tex]a_2=18\times\dfrac{2}{3} =12[/tex]
[tex]a_4=12\times\dfrac{2}{3} =8[/tex]
Thus the 4th term of the given geometric series is 8.
Learn more about the geometric series here;
https://brainly.com/question/1504226
