Question

What is r for the geometric series with s5=484, a1=4 and a5=324

Answer 1

Hello,

r=3



`a_(1)=4`

`a_(2)=4*r`

`a_(3)=4*r^2`

`a_(4)=4*r^3`

`a_(5)=4*r^4=324`
==>r=3 or r=-3
If r=-3 then
`s5_(3)=4*(1+r+r^2+r^3+r^4)=244`≠484 is rejected

if r=3 then
`s5_(3)=4*(1+r+r^2+r^3+r^4)=484`

Answer 2

r = 3

Explanation:

The geometric series is :

`\sum_(n=1)^(\infty) a_1 (r)^(n-1)`

a1=4 and a5 is the fifth element of the summation, that is:

`a_5 = a_1(r)^4 = 324`

`4(r)^4 = 324`

`(r)^4 = 81`

Either r = 3 or r = -3

s5 is the summation until the 5th element, that is:

`s_5 = a_1 (r) + a_1 (r)^1 + a_1 (r)^2 + a_1 (r)^3 + + a_1 (r)^4`

`s_5 = a_1 * (1 + r^1 + r^2 + r^3 + r^4)`

With r = 3

`s_5 = 4 * (1 + 3^1 + 3^2 + 3^3 + 3^4)`

`s_5 = 484`

With r = -3

`s_5 = 4 * (1 + (-3)^1 + (-3)^2 + (-3)^3 + (-3)^4)`

`s_5 = 244`

Therefore, r = 3