Please help me with 23

Solve for d

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}

rick has $1400 and sally has $1500

step-by-step explanation:

to find amount of money rick has after 10 years :

principal = $1000 , time = 10 years , rate = 4% annually

to find amount of money sally has after 10 years :

her friend gives to pay her $50 per year, therefore money she has after 10 years = 50 × 10 = $500

also, sally gets $1000 after 10 years,

therefore, total amount of money sally has after 10 years = 500 + 1000 = $1500

so, after 10 years : rick has $1400 and sally has $1500

95.03 is the answer. hope this !