T43 = T28 + 15(T28-T23)/(28-23)
= 24 + 15(24-16)/5 = 48
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?
7 answers
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.
81 is the correct answer, not 48
81
81
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*}
That is whyy the solution is 81.
2 answers