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

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 \(s5_{3}=4*(1+r+r^2+r^3+r^4)=484\)

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jbiain jbiain · Answer 2 · 2020-03-05T19:57:10-05:00

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