find the 8th term of a geometric sequence whose first term is -6 and whose third term is -8/3
Answers
Answer:
-(256/729) or 256/729
Step-by-step explanation:
Let, First term be a.
and common multiple be q in that geometric sequence.
Therefore 3rd term will be a*q^2 and 8th term will be a*q^7
According to the question ratio of third to first term is -(8/3) ÷ -(6) = 4/9
That is (a*q^2)/a = 4/9 => q^2 = 4/9 since q can both be positive and negative q = 2/3, -(2/3)
If q= 2/3 then 8th term, a*q^7 = -6*(2/3)^7 = - (2^8)/(3^6) = -256/729
If q= -(2/3) then 8th term, a*q^7 = -6*(-(2/3))^7 = (2^8)/(3^6) = 256/729
Therefore 8th term can be -256/729 or 256/729.
Answer:
GIVEN:
a1 = -6
a3 = -8/3
Step-by-step explanation:
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Solutions:
First, solve for common ratio
n = 3 (number of terms from a1 to a3)
an = a1 • r^n-1
a3 = a1 • r^3-1
-8/3 = -6r² *divide both sides by -6
(-8/3)/-6 = -6r²/-6
-8/3 x -? = r²
8/18 = r²
4/9 = r²
v4/9 = vr²
2/3 = r
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Next, Solve for the 8th term
n = 8
a8 = a1 • r^n-1
= -6 • ?^8-1
= -6 • ?7
= -6 (128/2187)
= -768/2187
a8 = -256/729