Question

How do you write a rule for the nth term of the geometric sequence and then find `a_5` given `a_3=-28, r=7/3`?

Answer 1

`-1372/9`

Explanation:

Let say the sequences are,
`a_1, a_2, a_3...a_n`, where,
`a_1 = a_1*r^0 =a_1*r^(1-1)`
`a_2 = a_1*r =a_1*r^(2-1)`
`a_3 = a_2*r = (a_1*r)*r = a_1*r^2 = (a_1*r)*r = a_1*r^(3-1)`
Therefore,
`T_n = a_1r^(n-1)`

given that `a_3 = -28 and r = 7/3`

`a_5 = a_1*r^(5-1) = a_1*r^4 = a_1*r^2*r^2`

`a_3=a_1*r^2` therefore,
`a_1*r^2*r^2 = a_3*r^2 = -28 * (7/3)^2 = -1372/9`

Answer 2

`a_n=-36/7(7/3)^(n-1),a_5=-1372/9`

Explanation:

The standard geometric sequence is.

`a,ar,ar^2,ar^3,......,ar^(n-1)`

where a is the first term, r is the common ratio and the nth term is

`• ar^(n-1)`

Each term in the sequence is obtained by multiplying the previous term by r, the common ratio.

`a_3=ar^2=-28`

`rArraxx(7/3)^2=-28`

`rArraxx49/9=-28`

`rArra=-28xx9/49=-36/7`

`rArra_n=-36/7(7/3)^(n-1)larrcolor(red)" nth term formula"`

`rArra_5=-36/7xx(7/3)^4=-1372/9`