Question

Write an explicit rule for t(n) of the geometric sequence below:

2, 3/2, 9/8, 27/32 .....

Answer 1

2* (3/4)=3/2

Answer 2

so thats the rule

Answer 3

no

Answer 4

one sec

Answer 5

\[a_{n}=ar^{n-1}\]is the general formula for geometric sequences. Here the "common ratio" is \[r=\frac{\frac{3}{2}}{2}=\frac{3}{4}\]The first term "a" is 2. So we have:\[t(n)=2(\frac{3}{4})^{n-1}\]where\[n=1,2,3...\]

Answer 6

ok

Answer 7

so will the rule be n(t)=1,2,3

Answer 8

no. The rule is,\[t(n)=2(\frac{3}{4})^{n-1}\]where,\[n=1,2,3...\]

Answer 9

so the rule is 2(3/4)n−1

Answer 10

(n-1) is the exponent on the (3/4) term. It is:\[(\frac{3}{4})^{n-1}\]NOT,\[\frac{3}{4}(n-1)\]

Answer 11

so it is (3/4)^n-1

Answer 12

For the (3/4) term, yes. look at my previous post for the whole equation.

Answer 13

so 2(3/4)^n-1

Answer 14

2(3/4)^{n-1}

Answer 15

Notice:\[2(\frac{3}{4})^{n-1}\neq 2(\frac{3}{4})^n-1\]

Answer 16

ok

Answer 17

2(3/4)^{n-1} it is like this

Answer 18

yep. So the first term, n=1, gives, 2(3/4)^{1-1}=2(3/4)^(0)=2, as expected. And the second term in the sequence is...n=2......
2(3/4)^{2-1}=3/2....etc...

Answer 19

ok i got it

Answer 20

cool