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# help pls

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$$Let g_1, g_2, g_3,\dots be a geometric sequence. If g_{23} = 16 and g_{28} = 24, what is g_{43}?$$

Jun 6, 2022

### 3+0 Answers

#1
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subtract 43 with 28, which is 15, and then multiply that by 1.6, and that is 24. after that, add 24 with 24 so g_43 = 48. Idk if this is right or wrong, I tried something else but the logic was wrong

Jun 6, 2022
#2
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By the property of geometric sequences, $$g_{43} = g_1 \times d^{42}$$

Note that $$g_{28} = g_{23} \times d^5$$. Substituting the given values, we have: $$28 = 16d^5$$, meaning $$d = \sqrt[5]{7 \over 4}$$.

Now, notice that $$g^{43} = g^{28} \times d^{15}$$

We know that $$\sqrt[5]{7 \over 4} = {7 \over 4}^3 = {343 \over 64}$$

This means that $$d^{43} = 28 \times {343 \over 64} = \color{brown}\boxed{150 {1 \over 16}}$$

Jun 6, 2022
#3
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I did not get either of these answers.

$$ar^{22}=16\qquad and \qquad ar^{27}=24\\~\\ \frac{ar^{27}}{ar^{22}}=\frac{24}{16}\\ r^5=\frac{3}{2}\\~\\ r=\left( \frac{3}{2} \right)^{1/5} \\~\\ ar^{22}=16\\~\\ a=16*\left( \frac{2}{3} \right)^{22/5} \\~\\ g_{43}=ar^{42}\\~\\ g_{43}=16*\left( \frac{2}{3} \right)^{22/5} \left(\left( \frac{3}{2} \right)^{1/5}\right)^{42}\\~\\ \displaystyle g_{43}=\frac{2^4*2^{22/5}*3^{42/5}}{3^{22/5}*2^{42/5}}\\~\\ \displaystyle g_{43}=\frac{2^4*3^{20/5}}{2^{20/5}}\\~\\ \displaystyle g_{43}=\frac{2^4*3^{4}}{2^4}\\~\\ \displaystyle g_{43}=3^4\\~\\ \displaystyle g_{43}=81\\~\\$$

LaTex

ar^{22}=16\qquad and \qquad ar^{27}=24\\~\\
\frac{ar^{27}}{ar^{22}}=\frac{24}{16}\\
r^5=\frac{3}{2}\\~\\
r=\left(  \frac{3}{2} \right)^{1/5} \\~\\
ar^{22}=16\\~\\
a=16*\left(  \frac{2}{3} \right)^{22/5} \\~\\
g_{43}=ar^{42}\\~\\
g_{43}=16*\left(  \frac{2}{3} \right)^{22/5} \left(\left(  \frac{3}{2} \right)^{1/5}\right)^{42}\\~\\

\displaystyle g_{43}=\frac{2^4*2^{22/5}*3^{42/5}}{3^{22/5}*2^{42/5}}\\~\\
\displaystyle g_{43}=\frac{2^4*3^{20/5}}{2^{20/5}}\\~\\
\displaystyle g_{43}=\frac{2^4*3^{4}}{2^4}\\~\\
\displaystyle g_{43}=3^4\\~\\
\displaystyle g_{43}=81\\~\\

Jun 7, 2022