In a geometric progression ,the sum of 3rd and 4th term is 108. if the sum of the 4th and 5th term is 324, determine the: common ratio, first term and the 12th term.

see explanation

Step-by-step explanation:

The terms of a geometric progression are

a, ar, ar², ar³, ...... , a

Thus the sum of the third and fourth is

ar² + ar³ = 108 → (1)

The sum of the fourth and fifth is

ar³ + a = 324 → (2)

Factorise both equations

ar²(1 + r) = 108 → (3)

ar³(1 + r) = 324 → (4)

Divide (4) by (3)

=

Cancelling (1 + r) , a and r, gives

r = 3

Substitute r = 3 into (3) and solve for a

9a(4) = 108

36a = 108 ( divide both sides by 36 )

a = 3

The n th term is

= a with a = 3 and r = 3, hence

= 3 = 531441

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}