How do you write a rule for the nth term of the geometric term and then find $a_6$ given $r=-1/2, a_3=8$ ?

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Answer 1 · 2017-06-14T10:52:03

$a_n=a_1*r^(n-1) and a_6= -1$

Rule of finding n th term of G.P is $a_n=a_1*r^(n-1)$ .where $a_1,r,n$
are 1st term, common ratio and required term respectively.

$a_3=8 , r=-1/2, a_6= ?$

$a_3= a_1* (-1/2)^(3-1) or 8= a_1* (-1/2)^2 or 8 = a_1*1/4$ or

$a_1=32 ; :. a_6= a_1* (-1/2)^(6-1) or a_6 = 32* (-1/2)^5$ or

$a_6 = 32* (-1/32) = -1$

$a_6= -1$ [Ans]

Answer 2 · 2017-06-14T11:01:43

$a_n=32(-1/2)^(n-1), and, a_6=-1.$

In the Usual Notation for Geometric Sequence,

we are given, $r=-1/2, and, a_3=8....(1).$

Knowing that, $a_n=a_1*r^(n-1)," we have, using "(1),$

$8=a_3=a_1*(-1/2)^(3-1), or, a_1=32.$

$:. a_n=32(-1/2)^(n-1).$

$:. a_6=32(-1/2)^5=-1.$

How do you write a rule for the nth term of the geometric term and then find a_6a_6 given r=-1/2, a_3=8r=-1/2, a_3=8?