Asked by Speedo
The 23rd term in a certain geometric sequence is 16 and the 28th term in the sequence is 24. What is the 43rd term?
What is the answer?
What is the answer?
Answers
Answered by
Steve
T43 = T28 + 15(T28-T23)/(28-23)
= 24 + 15(24-16)/5 = 48
= 24 + 15(24-16)/5 = 48
Answered by
Anonymous
I don't understand why so many people are getting 48, but that is not the answer. We know that in a geometric sequence, you multiply each term by a number to get your next term. So we know that we multiply our 23rd term by some number x 5 times to get our 28th number. So we know that 16 * x^5 = 24. All I did here was plug in the numbers. Therefore x^5 = 24/16 = 3/2, so x equals the fifth root of 3/2. We know that 24 * (fifth root of 3/2)^15 = T43 (T43 = 43rd term) . Simplifying this, we get T43 = 81.
Answered by
Girl
81 is the correct answer, not 48
Answered by
Me
81
Answered by
oh
81
Answered by
TheAopsMachinist
Let $r$ be the common ratio of the geometric sequence. To get from the $23^\text{rd}$ term of the sequence to the $28^\text{th}$ term, we start with the $23^{\text{rd}}$ term and multiply by $r$ five times. Therefore, we have
\[16r^5 = 24,\]so $r^5 = \frac{24}{16} = \frac32$. To get from the $28^\text{th}$ term to the $43^\text{rd}$ term, we start with the $28^\text{th}$ term and multiply by $r$ fifteen times. So, the $43^\text{rd}$ term is
\begin{align*}
24r^{15} &= 24\left(r^5\right)^3 \\
&= 24\left(\frac32\right)^3 \\
&= 24\cdot \frac{27}{8} \\
&= \boxed{81}.
\end{align*}
\[16r^5 = 24,\]so $r^5 = \frac{24}{16} = \frac32$. To get from the $28^\text{th}$ term to the $43^\text{rd}$ term, we start with the $28^\text{th}$ term and multiply by $r$ fifteen times. So, the $43^\text{rd}$ term is
\begin{align*}
24r^{15} &= 24\left(r^5\right)^3 \\
&= 24\left(\frac32\right)^3 \\
&= 24\cdot \frac{27}{8} \\
&= \boxed{81}.
\end{align*}
Answered by
TheAopsMachinist
That is whyy the solution is 81.
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