Question

The $23 rd term in a certain geometric sequence is 16 and the $28th term in the sequence is 24. What is the $43 rd term?

Answer 1

Answer:

81

Step-by-step explanation:

Geometric sequence formula:  $a_n=ar^{n-1}$

Given:

• $a_{23}=ar^{22}=16$
• $a_{28}=ar^{27}=24$

Find common ratio (r):

$\begin{aligned}\implies \dfrac{a_{28}}{a_{23}} =\dfrac{ar^{27}}{ar^{22}} & =\dfrac{24}{16}\\ \implies r^5 & =\dfrac32\\ \implies r & = \sqrt[5]{\frac32} \end{aligned}$

Find initial term (a):

$\implies ar^{22}=16$

$\implies a(\sqrt[5]{\frac32} )^{22}=16$

$\implies a(\frac32} )^{\frac{22}{5}}=16$

$\implies a=\dfrac{16}{(\frac32)^{\frac{22}{5}}}$

Find the 43rd term:

$\implies a_{43}=ar^{42}$

$\implies a_{43}= \left(\dfrac{16}{(\frac32)^{\frac{22}{5}}}\right)(\sqrt[5]{\frac32} )^{42}$

$\implies a_{43}=16 \cdot (\frac32)^{-\frac{22}{5}} \cdot (\frac32)^{\frac{42}{5}}$

$\implies a_{43}=16 \cdot (\frac32)^4$

$\implies a_{43}=16 \cdot \left(\dfrac{3^4}{2^4}\right)$

$\implies a_{43}=16 \cdot \left(\dfrac{81}{16}\right)$

$\implies a_{43}=81$