How do you write a rule for the nth term of the geometric term given the two terms $a_4=-8/9, a_7=-64/243$ ?

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Answer 1 · 2017-02-28T11:35:19

$a_n = -2^(n-1)/3^(n-2)$

Suppose using standard notation for a GP sequence that the first term is $a$ and the common ratio is $r$ , the first few terms of the sequence are:

$( a, ar, ar^2, ar^3, ar^4 , ... }$

Assuming that the first term is $a_1$ , then;

$a_1 = a$
$a_2 = ar$
$a_3 = ar^2$
$vdots$
$a_n = ar^(n-1)$

Then $a_4=-8/9 \ \ \ => ar^3 = -8/9 \ \ \ \ \ \ .....[1]$
And, $a_7 = -64/243 => ar^6 = -64/243 \ \ \ .....[2]$

$Eq [2] divide Eq[1]$ gives;

$\ \ (ar^6)/(ar^3) = (-64/243)/(-8/9)$
$:. r^3 = 64/243*9/8$
$:. r^3 = 8/27$
$:. \ \ r = 2/3$

Subs $r^3 = 8/27$ into $Eq[1]$ gives:

$a*8/27 = -8/9$
$:. a = -8/9*27/8$
$\ \ \ \ \ \ \ = -3$

And so the terms form a GP with $a=-3$ , $r=2/3$

So the $n^(th)$ term is given by:

$a_n = ar^(n-1)$
$\ \ \ \ = -3(2/3)^(n-1)$
$\ \ \ \ = -2^(n-1)/3^(n-2)$

Check

$n=4 => a_4 = -2^3/3^2 = -8/9$
$n=7 => a_7 = -2^6/3^5 = -64/243$

How do you write a rule for the nth term of the geometric term given the two terms a_4=-8/9, a_7=-64/243a_4=-8/9, a_7=-64/243?