# Common ratio Math help please!?

Find the common ratio for each of the following geometric series. Determine whether a sum to infinity exists. Find the sum where it exists.
a) 9 + 3 + 1 ....
r=?
S=?

b) 7-1 + 1/7 - .....
r=?
S=?

c) -4 +6 - 9 + ....
r=?
S=?

a) 9 + 3 + 1 ....
r=?
S=?

geometric series eqations is a+ar+ar^2+ar^3+----------- and so on

Here first term is a

And r = t2/t1

[Note : t2 means second term in geometric series , t1 means first term in geometric series ]

9 + 3 + 1 ....

Here a = 9

And  r = 3/9

Simplfy

r = 1/3

[Formula : geometric series eqations in S(infinite) = a / (1-r )]

Substitute a = 9 and r = 1/3 value in above formula

S = 9/( 1-(1/3) )

S = 9/( (3-1)/3)                          here 1 ,3 LCM is 3

S = 9/ ( 2/3 )

Multiply each side by '2/3'.

S(2/3) = [9/ ( 2/3 )] ( 2/3 )

Simplfy

S(2/3) = 9

Multiply each side by '3'

S( 2/3 )(3) = 9( 3 )

Simplfy

S ( 2 ) = 27

Divide each side by 2.

S (2)/2 = 27/2

Simplfy

S = 27/2

There fore r = 1/3 and S = 27/2

In geometric series, S(infinite) = a / (1-r )

Substitute a = 9 and r = 1/3 value in above formula

S = 9/( 1-(1/3) )

S = 9/( (3-1)/3)                          here 1 ,3 LCM is 3

S = 9/ ( 2/3 )

S = (9*3) / 2

S = 27 / 2

b) 7-1 + 1/7 - .....
r=?
S=?

geometric series eqations is a+ar+ar^2+ar^3+----------- and so on

Here first term is a

And r = t2/t1

[Note : t2 means second term in geometric series , t1 means first term in geometric series ]

7-1 + 1/7 - .....

Here a = 7

And  r = -1/7

[Formula : geometric series eqations in S(infinite) = a / (1-r )]

Substitute a = 7 and r = -1/7 value in above formula

S = 7/( 1-(-1/7) )

S = 7/( (7+1)/7)                          here 1 ,7 LCM is 7

S = 7/ ( 8/7 )

Multiply each side by '8/7'.

S(8/7) = [7/ ( 8/7 )] ( 8/7 )

Simplfy

S(8/7) = 7

Multiply each side by '7'

S( 8/7 )(7) = 7( 7 )

Simplfy

S ( 8 ) = 49

Divide each side by 8.

S (8)/8 = 49/8

Simplfy

S = 49/8

There fore r = -1/7 and S = 49/8

c) -4 +6 - 9 + ....
r=?
S=?

geometric series eqations is a+ar+ar^2+ar^3+----------- and so on

Here first term is a

And r = t2/t1

[Note : t2 means second term in geometric series , t1 means first term in geometric series ]

-4 +6 - 9 + ....

Here a = -4

And  r = 6/(-4)

Simplfy

r = -3/2

[Formula : geometric series eqations in S(infinite) = a / (1-r )]

Substitute a = -4 and r = -3/2 value in above formula

S = -4/( 1-(-3/2 )

Multiply two negative signs are positive

S = -4/( 1+ (3/2) )

S = -4/( (2+3)/2)                          here 1 ,2 LCM is 2

S = -4 / ( 5/2 )

Multiply each side by '5/2'.

S(5/2) = [-4/ ( 5/2 )] ( 5/2 )

Simplfy

S(5/2) = -4

Multiply each side by '2'

S( 5/2 )(2) = -4( 2 )

Simplfy

S ( 5 ) = -8

Divide each side by 5.

S (5)/5 = -8/5

Simplfy

S = -8/5

There fore r = -3/2 and S = - 8/5

a) 9 + 3 + 1 ....

In geometric series, common ratio r = second term/ first term = a₂ / a₁.

Therefore r = 3/9 = 1/3.

In geometric series, sum of n terms S(infinite) = first term / (1 - common ratio).

Therefore S = a / 1 - r

= 9 / (1 - 1/3)

= 9 / (3-1)/3

= 9*3 / 2

= 27/2.

Therefore r = 1/3 and S = 27/2.

b) 7-1 + 1/7 - ......

In geometric series, common ratio r = second term/ first term = a₂ / a₁.

Therefore r = -1/7.

In geometric series, sum of n terms S(infinite) = first term / (1 - common ratio).

Therefore S = a / 1 - r

= 7 / (1 - (-1/7))

= 7 / (1 + 1/7)

= 7 / (7+1)/7

= 7*7 / 8

= 49/8.

Therefore r = -1/7 and S = 49/8.

c) -4 +6 - 9 + ....

In geometric series, common ratio r = second term/ first term = a₂ / a₁.

Therefore r = 6/-4 = -3/2.

In geometric series, sum of n terms S(infinite) = first term / (1 - common ratio).

Therefore S = a / 1 - r

= -4 / (1 - (-3/2))

= -4 / (1 + 3/2)

= -4 / (2+3)/2

= -4*2 / 5

= -8/5.

Therefore r = -3/2 and S = -8/5.