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

a*r^0+a*r^1+a*r^2 1 2 3 notice a patter?

your a is 1024

huh?

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 :)

whats r? haha(:

r=-1/4 so 1024*(-1/4)^5

-1?

it is actually -1/4

your r

-1 is your answer yes :D lol sorry

See you can do it ;)

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(:

sure :)

thank youu!!(:

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 ;)

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

.008?

YES

thank you!!(: your soo helpful(:

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

a*r^0+a*r^1+a*r^2 1 2 3 notice how the r^n is just -1 the nth term

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

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