Find the sixth term of a geometric sequence with t₅ = 24 and t₈ = 3.

Solution:

Given, t₅ = 24 and t₈ = 3

The n-th term of the geometric sequence is given by

\(t_{n}=t(r)^{(n-1)}\)

\(\\t_{5}=t(r)^{(5-1)}\\24=tr^{4}\)

t = 24/r⁴

\(\\t_{8}=t(r)^{(8-1)}\\3=tr^{7}\)

Now, 3 = (24/r⁴)(r⁷)

3 = 24r³

r³ = 3/24

r³ = 1/8

Taking cube root,

r = 1/2

Now, 24 = a(1/2)⁴

24 = a(1/16)

a = 384

The sixth term of the sequence, \(\\t_{6}=tr^{(6-1)}\\t_{6}=384(1/2)^{5})\\t_{6}=384(1/32)\)

\(t_{6}=12\)

Therefore, the sixth term is 12.

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Find the sixth term of a geometric sequence with t₅ = 24 and t₈ = 3.

Summary:

The sixth term of a geometric sequence with t₅ = 24 and t₈ = 3 is 12.