# What is the 7th term of the geometric sequence where a_{1} = 256 and a_{3} = 16?

**Solution:**

The nth term of the geometric sequence is given by:

a_{n} = a · r^{n - 1},

Where a is the first term and r is the common ratio respectively.

Given: a_{1} = a = 256 and a_{3} = 16

We know, a_{n} = a · r^{n - 1}

⇒ a_{3} = a. r^{3 - 1}

⇒ 16 = 256 × r^{2}

⇒ r^{2} = 16/256

⇒ r^{2 }= 1/16

⇒ r = ±1/4

Now, a_{7} = a. r^{7 - 1} = a . r^{6}

⇒ a_{7} = 256 × (1/4)^{6}

⇒ a_{7 }= 256/4096

⇒ a_{7 } = 1/16

If r = -1/4, then

⇒ a_{7} = 256 × (-1/4)6

⇒ a_{7 }= 256/4096

⇒ a_{7} = 1/16

Therefore, the 7th term of the geometric sequence a_{7} is 1/16.

## What is the 7th term of the geometric sequence where a_{1} = 256 and a_{3} = 16?

**Summary:**

The 7th term of the geometric sequence where a_{1} = 256 and a_{3} = 16 is 1/16.