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

Answer:

r = 3

Step-by-step 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:

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

$s_5 = a_1 \times (1 + r^1 + r^2 + r^3 + r^4)$

With r = 3

$s_5 = 4 \times (1 + 3^1 + 3^2 + 3^3 + 3^4)$

$s_5 = 484$

With r = -3

$s_5 = 4 \times (1 + (-3)^1 + (-3)^2 + (-3)^3 + (-3)^4)$

$s_5 = 244$

Therefore, r = 3