How do you write the expression for the nth term of the geometric sequence $a_{1} = 16, a_{4} = \frac{27}{4}, n = 3$ $a_1=16, a_4=27/4, n=3$ ?

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Answer 1 · 2018-02-23T12:56:08

$n$ $n$ th term is $a_{1} \cdot r^{n - 1}$ $a_1*r^(n-1)$ and $a_{3} = 16$ $a_3=16$

In Geometric progression $n$ $n$ th term is $a_{1} \cdot r^{n - 1}$ $a_1*r^(n-1)$ where

$a_{1}$ $a_1$ is first term and $r$ $r$ is the common ratio.

$a_{1} = 16, a_{4} = \frac{27}{4}$ $a_1=16 , a_4=27/4$

$:. a_4=16*r^(4-1) or a_4= 16r^3 = 27/4 or r^3= 27/64$ or

$r= root(3)(27/64)=3/4$ . When $n=3 :. a_3= 16* (3/4)^(3-1)$

or $a_3= 16* (3/4)^(3-1)=16*(3/4)^2=9$

How do you write the expression for the nth term of the geometric sequence a_1=16, a_4=27/4, n=3a1=16,a4=274,n=3a_1=16, a_4=27/4, n=3?