Question

how do I solve, What is the 6th term of the geometric sequence where a1 = 1,024 and a4 = –16

Answer 1

a*r^0+a*r^1+a*r^2
1 2 3

notice a patter?

Answer 2

your a is 1024

Answer 3

huh?

Answer 4

a4 will be a*r^3 =-16

1024*r^3=-16 SFr

and plug r back into 6th term which will be

a*r^5

1024*(r you found above)^5 :)

Answer 5

whats r? haha(:

Answer 6

r=-1/4

so
1024*(-1/4)^5

Answer 7

-1?

Answer 8

it is actually -1/4

Answer 9

your r

Answer 10

-1 is your answer yes :D
lol sorry

Answer 11

See you can do it ;)

Answer 12

haha thanks so I know its the same type of problems but they confuse me so, can you help me with What is the 8th term of the geometric sequence where a1 = 625 and a3 = 25? pleaseee(:

Answer 13

sure :)

Answer 14

thank youu!!(:

Answer 15

a*r^0+a*r^1+a*r^2
1 2 3

625 is your a

625*r^2=25 sfr

then

625*(r you solved for)^7

;)

Answer 16

r is 1/5 :) can you take it from here?

Answer 17

.008?

Answer 18

YES

Answer 19

thank you!!(: your soo helpful(:

Answer 20

So long as you know how to do it :D and notice the pattern

Answer 21

a*r^0+a*r^1+a*r^2
1 2 3


notice how the r^n is just -1 the nth term

Answer 22

r^0 r^1 r^2 r^3

n1 n2 n3 n4


ooh ni is always a your starting a.

cause a*r^0 =a

r^0 = 1 so it is just a * 1= a