Question

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

Answer 1

Answer:

r = 2

Step-by-step explanation:

nth term of a geometric sequence = $ar^{n - 1}$.

Where,

a = first term, r = common ratio,

We are given that, the 11th term is 32 times larger than the 6th term of the geometric sequence. Thus:

the 6th term can be expressed as $ar^{n - 1} = ar^{6 - 1}$

$a_6 = ar^{5}$

The 11th term is expressed as $a_11 = ar^{10}$

Since the 11th term is 32 times larger than the 6th term, therefore, the following equation can be derived to find the common ratio, r, of the sequence:

$a_11 = 32*a_6$

$a_11 = ar^{10}$

$a_6 = ar^{5}$, therefore:

$ar^{10} = 32*ar^{5}$

Divide both sides by ar⁵

$\frac{ar^{10}}{ar^5} = \frac{32*ar^{5}}{ar^5}$

$r^5 = 32$

$r^5 = 2^5$

$r = 2$

Common ratio = 2