Question

If the 11th of a geometric sequence is 32 times larger than the 6th term, then what is the common ratio of the sequence?

Answer 1

Answer:

2

Step-by-step explanation:

Hello, let's note $(a_n)$ this sequence and r the ratio, we can write

$a_n=a_0r^n$

With $a_0$ different from 0 and then

$a_{11}=32a_6<=>a_or^{11}=32a_0r^6<=>r^{11-6}=32=2^5<=>r^5=2^5<=>r=2$

Thanks