Question

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?

Answer 1

2

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