Mathematics OpenStudy (anonymous):

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

a*r^0+a*r^1+a*r^2 1 2 3 notice a patter? OpenStudy (anonymous): OpenStudy (anonymous):

huh? OpenStudy (anonymous):

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 :) OpenStudy (anonymous):

whats r? haha(: OpenStudy (anonymous):

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

-1? OpenStudy (anonymous):

it is actually -1/4 OpenStudy (anonymous): OpenStudy (anonymous): OpenStudy (anonymous):

See you can do it ;) OpenStudy (anonymous):

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(: OpenStudy (anonymous):

sure :) OpenStudy (anonymous):

thank youu!!(: OpenStudy (anonymous):

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 ;) OpenStudy (anonymous):

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

.008? OpenStudy (anonymous):

YES OpenStudy (anonymous): OpenStudy (anonymous):

So long as you know how to do it :D and notice the pattern OpenStudy (anonymous):

a*r^0+a*r^1+a*r^2 1 2 3 notice how the r^n is just -1 the nth term OpenStudy (anonymous):

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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