Question

Find an equation for the nth term of a geometric sequence where the second and fifth terms are -21 and 567, respectively.

Answer 1

I know An=A1+(n−1)d, but I don't know how to find difference or the first term

Answer 2

no, for geometric series
\[r=\frac{ a _{n} }{ a _{n-1} }\]
\[t _{n}=a r ^{n-1}\]

Answer 3

Oh, okay then i really have no clue what I'm doing. I'm trying to read the online lesson but it makes no sense to me, could you explain?

Answer 4

put n=2,
n=5
divide and find r
find a
find \[t _{n}\]

Answer 5

to start with let a be the first term and r be common ratio of the given G.P

Answer 6

so tn=(-21)r (2)−1? I dont really understand

Answer 7

\[t _{2}=ar ^{2-1}=ar=-21\]
\[t _{5}=?\]

Answer 8

oh, okay.. t5= -21

Answer 9

no,\[t _{2}=ar=-21 ...(1)\]
\[t _{5}=ar ^{5-1}=567\]
\[ar^4=567 ...(2)\]
divide (2) by (1)

Answer 10

oh sorry, i meant t3=-21
so ar4= 1,134?? i feel im not understanding

Answer 11

\[\frac{ ar^4 }{ ar }=\frac{ 567 }{ -21 }\]
\[r^3=-27=\left( -3 \right)^3,r=-3\]
ar=-21
a*(-3)=-21
\[a=\frac{ -21 }{ -3 }=7\]
\[t _{n}=?\]

Answer 12

plug the values of a and r and find \[t _{n}\]

Answer 13

tn= (7)(-3)^n−1

Answer 14

idk why its doing that, i put n-1

Answer 15

\[t _{n}=7\left( -3 \right)^{n-1}\]

Answer 16

i got it! thank you so much for taking your time!(:

Answer 17

i am leaving and good night.