Answer:
Step-by-step explanation:
Given two series A and B;
Series A: 10 + 4 +8/5 + 16/25 + 13/125 + ...
Series B: 1/5 + 3/5 + 9/5 + 27/5 + 81/5 +...
Both series are geometric series since they have the same common ratio. Given the geometric series T₁, T₂, T₃, T₄..., the common ratio r is expressed as [tex]\frac{T_2}{T_1} =\frac{T_3}{T_2} =\frac{T_4}{T_3} = r[/tex].
a) For series A, Given T₁ = 10, T₂ = 4, T₃ = 8/5 and T₄ = 13/125
Its common ratio is expressed as [tex]\frac{4}{10} =\frac{8/5}{4} =\frac{13/125}{8/5} = 2/5[/tex]
Hence the common ratio of series A is 2/5
The sum to infinity of the geometric series will be expressed as [tex]s_\infty = \dfrac{a}{1-r}[/tex]
Note that the sum to infifnity is used since the series is tending to infinity.
a is the first term if the series = 10
r - 2/5
On substituting this values into the formula;
[tex]s_\infty = \dfrac{10}{1-2/5} \\\\s_\infty = \dfrac{10}{3/5} \\\\s_\infty = 10* 5/3\\s_\infty = 50/3[/tex]
Hence the sum of the series is 50/3
b) For the series B;
Series B: 1/5 + 3/5 + 9/5 + 27/5 + 81/5 +...
The common ratio of the series is expressed as shown;
[tex]\frac{3/5}{1/5} =\frac{9/5}{3/5}= \frac{27/5}{9/5 } = r\\ r = (3/5 * 5/1) = 9/5*5/3 = 27/5*5/9 = 3[/tex]
Hence its common ratio is 3.
Sun of the series is expressed similarly as:
[tex]s_\infty = \frac{a}{1-r}\\ s_\infty = \frac{1/5}{1-3}\\\\s_\infty = \frac{1/5}{-2}\\\\[/tex]
[tex]s_\infty = 1/5 * -1/2\\s_\infty = -1/10[/tex]
Hence the sum of series B is -1/10
