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

2

Step-by-step explanation:

Given that:

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

The nth term of a geometric sequence is given by:

An = A*r^(n-1)

Where An = the nth term ; r = common ratio and n = term number ; A = first term

The 11th term:

A6 = A*r^(6 - 1) = Ar^5

A11 = 32A6 = 32Ar^5

A11 = Ar^(11 - 1) = Ar^10

Hence,

Ar^10 = 32Ar^5

r^(10 - 5) = 32

r^5 = 2^5

Hence, r = 2

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.