Question

Determine the common ratio of the infinite geometric series

25) a1 = 1, S = 1.25

26) a1 = 96, S = 64

27) a1 = −4, S = −16/5

28) a1 = 1, S = 2.5

Answer 1

25)

Given S  =  1.25 and a  =  1

The sum of the infinite series in GP is

S = a / (1 - r)

Substitute S = 1.25 and a = 1 in above equation

1.25  =  1 / (1 - r)

1.25(1 - r)  =  1

1.25 - 1.25r  =  1

1.25 - 1  =  1.25r

0.25  =  1.25r

r  =  0.25/1.25

r  =  0.2

 

26)

Given S  =  64 and a  =  96

The sum of the infinite series in GP is

S = a / (1 - r)

Substitute S = 64 and a = 96 in above equation

64  =  96 / (1 - r)

64(1 - r)  =  96

64 - 64r  =  96

64 - 96  =  64r

-32  =  64r

r  =  (-32)/64

r  =  - 0.5

 

27)

Given S  =  -16/5 and a  =  -4

The sum of the infinite series in GP is

S = a / (1 - r)

Substitute S = -16/5 and a = -4 in above equation

-16/5  =  (-4) / (1 - r)

-16(1 - r)  =  (-4)5

-16 + 16r  =  - 20

16r  =  - 20 + 16

16r  =  - 4

r  =  (-4)/16

r  =  - 0.25

 

28)

Given S  =  2.5 and a  =  1

The sum of the infinite series in GP is

S = a / (1 - r)

Substitute S = 2.5 and a = 1 in above equation

2.5  =  1 / (1 - r)

2.5(1 - r)  =  1

2.5 - 2.5r  =  1

2.5 - 1  =  2.5r

1.5  =  2.5r

r  =  1.5/2.5

r  =  0.6