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The first term of a geometric sequence is 729, and the 7th term is 64. What is the positive, real value for the 5th term?

Guest May 8, 2018
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 #1
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[64/729]^1 / (N - 1)

[64/729]^1/6 = 2/3 Common ratio

5th term =F x R(N - 1)

              =729 x (2/3)^4

              =729 x 16/81

              = 144 - the fifth term of the GS.

Guest May 8, 2018
 #2
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The first term of a geometric sequence is 729, and the 7th term is 64.

What is the positive, real value for the 5th term?

 

\(\text{GP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $a =$ term }\\ \text{ $i,j,k = $ indices}\\\\ \large{ a_i^{j-k} \times a_j^{k-i} \times a_k^{i-j} = 1 } \\\\ & i = 1 & a_1 = 729 \\ & j = 7 & a_7 = 64 \\ & k = 5 & a_5 =\ ? \\ & j-k = 7-5 = 2 \\ & k-i = 5-1 = 4 \\ & i-j = 1-7 = -6 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline a_1^{2} \times a_7^{4} \times a_5^{-6} &=& 1 \quad & | \quad a_1 = 729 \quad a_7 = 64 \\ 729^{2} \times 64^{4} \times a_5^{-6} &=& 1 \\ 729^{2} \times 64^{4} &=& a_5^{6} \quad & | \quad ()^{\frac16} \\ 729^{\frac26} \times 64^{\frac46} &=& a_5^{\frac66} \\ 729^{\frac13} \times 64^{\frac23} &=& a_5 \\ 9 \times 16 &=& a_5 \\ \mathbf{a_5} & \mathbf{=} & \mathbf{144} \\ \hline \end{array}\)

 

laugh

heureka  May 9, 2018

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