Pick one of the remaining choices and explain on the lines below, why thatchoice contains two expressions that are not equivalent for any value of r.

If two expressions have the same coefficients of the variable, then the variable can be any number

For the first choice

3(3r + 3) = 9r + 9 === 1st expression

6r + 6 2nd expression

If we equate them

9r + 9 = 6r + 6

If we solve the equation we will find just one value of r

9r + 9 - 6r = 6r - 6r + 6

3r + 9 = 6

3r + 9 - 9 = 6 - 9

3r = -3

r = -1

But in the other choices, the two expressions have the same coefficients of r

Then when we equate them they will cancel each other

Then r can be any value

Because it does not affect the equation

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}