Question

What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?

Answer 1

Answer:

The sixth term of the geometric sequence is 4.

Step-by-step explanation:

Given : Geometric sequence term are $a_1=-4096$ and  $a_4=64$    

To find : What is the 6th term of geometric sequence ?

Solution :

$a_1=-4096$

We know, fourth term is $a_4=ar^3$

$a_4=64$    

So, $64=(-4096)r^3$

$\frac{64}{(-4096)}=r^3$

$-0.015625=r^3$

$r=(-0.015625)^{\frac{1}{3}}$

The sixth term is  $a_6=ar^5$

Substitute,

$a_6=(-4096)((-0.015625)^{\frac{1}{3}})^5$

$a_6=-4096((-0.015625)^{\frac{5}{3}})$

$a_6=-4096(-0.0009765625)$

$a_6=4$

Therefore, The sixth term of the geometric sequence is 4.

Answer 2

$\bf \begin{array}{llccll} term&value\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ a_1&-4096\\ a_2&-4096r\\ a_3&-4096rr\\ a_4&-4096rrr\\ &-4096r^3\\ &64 \end{array}\implies -4096r^3=64 \\\\\\ r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}} \\\\\\ r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\ -------------------------------$

$\bf n^{th}\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ r=-\frac{1}{4}\\ a_1=-4096\\ n=6 \end{cases} \\\\\\ a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5$