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Complete the statements about 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 +...

Series (a or b) has an r value of (blank) where 0<|r|<1. So, we can find the sum of the series. The sum of the series is (blank) ​

User Ehsan Kia
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Answer:

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
(T_2)/(T_1) =(T_3)/(T_2) =(T_4)/(T_3) = r.

a) For series A, Given T₁ = 10, T₂ = 4, T₃ = 8/5 and T₄ = 13/125

Its common ratio is expressed as
(4)/(10) =(8/5)/(4) =(13/125)/(8/5) = 2/5

Hence the common ratio of series A is 2/5

The sum to infinity of the geometric series will be expressed as
s_\infty = (a)/(1-r)

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;


s_\infty = (10)/(1-2/5) \\\\s_\infty = (10)/(3/5) \\\\s_\infty = 10* 5/3\\s_\infty = 50/3

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;


(3/5)/(1/5) =(9/5)/(3/5)= (27/5)/(9/5 ) = r\\ r = (3/5 * 5/1) = 9/5*5/3 = 27/5*5/9 = 3

Hence its common ratio is 3.

Sun of the series is expressed similarly as:


s_\infty = (a)/(1-r)\\ s_\infty = (1/5)/(1-3)\\\\s_\infty = (1/5)/(-2)\\\\


s_\infty = 1/5 * -1/2\\s_\infty = -1/10

Hence the sum of series B is -1/10

User Amit Rastogi
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