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# Help!

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We have two geometric sequences of positive real numbers:

6,a,b and 1/b,a,54

Solve for a

Guest May 7, 2018
#1
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We have two geometric sequences of positive real numbers:

6,a,b and 1/b,a,54

Solve for a

$$\text{Let a_n = a_1q^{n-1}} \\ \text{Let a_{n-1} = a_1q^{n-2}} \\ \text{Let a_{n+1} = a_1q^{n}}$$

$$\begin{array}{|rcll|} \hline a_{n-1}a_{n+1} &=& a_1q^{n-2}a_1q^{n} \\ &=& a_1^2q^{n-2}q^{n} \\ &=& a_1^2q^{n+n-2} \\ &=& a_1^2q^{2n-2} \\ &=& a_1^2q^{2(n-1)} \\ &=& a_1^2(q^{n-1})^2 \\\\ \sqrt{a_{n-1}a_{n+1}} &=& a_1 (q^{n-1}) \\ &=& a_n \\ \hline \end{array}$$

Formula:
$$\begin{array}{|rcll|} \hline a_n = \sqrt{a_{n-1}a_{n+1}} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1) & a &=& \sqrt{6b} \\ (2) & a &=& \sqrt{\frac1b\cdot 54} \\\\ & a=\sqrt{6b} &=& \sqrt{\frac1b\cdot 54} \\ & 6b &=& \frac1b\cdot 54 \\ & b^2 &=& \frac{54}{6} \\ & b^2 &=& 9 \\ & \mathbf{ b } & \mathbf{=} & \mathbf{3} \\ \hline \end{array}$$

$$a=\ ?$$

$$\begin{array}{|rcll|} \hline a &=& \sqrt{6b} \quad & | \quad b= 3 \\ a &=& \sqrt{6\cdot 3} \\ a &=& \sqrt{2\cdot 3^2} \\ \mathbf{a} &\mathbf{=}& \mathbf{3\sqrt{2}} \\ \hline \end{array}$$

heureka  May 7, 2018