Ineed plz i don not understand this problem

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}

answer: 0.1,0.85

step-by-step explanation:

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -