Question

What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?

Answer 1

`\large\boxed{\text{sixth term is equal to}\ (1)/(1024)}`

Explanation:

The explicit formula for a geometric sequence:

`a_n=a_1r^(n-1)`

`a_n` - n-th term

`a_1` - first term

`r` - common ratio

`r=(a_2)/(a_1)=(a_3)/(a_2)=...=(a_n)/(a_(n-1))`

We have

`a_1=-1,\ a_2=(1)/(4),\ a_3=-(1)/(6),\ ...`

The common ratio:

`r=((1)/(4))/(-1)=-(1)/(4)\\\\r=(-(1)/(6))/((1)/(4))=-(1)/(6)\cdot(4)/(1)=-(2)/(3)\\eq-(1)/(4)`

It's not a geometric sequence.

If
`a_3=-(1)/(16)` then the common ratio is
`r=(-(1)/(16))/((1)/(4))=-(1)/(16)\cdot(4)/(1)=-(1)/(4)`

Put to the explicit formula:

`a_n=-1\left(-(1)/(4)\right)^(n-1)`

Put
`a_n=(1)/(1024)` and solve for n :

`-1\left(-(1)/(4)\right)^(n-1)=(1)/(1024)\qquad\text{use}\ a^n:a^m=a^(n-m)\\\\-\left(-(1)/(4)\right)^n:\left(-(1)/(4)\right)^1=(1)/(1024)\\\\-\left(-(1)/(4)\right)^n\cdot(-4)=(1)/(1024)\\\\(4)\left(-(1)/(4)\right)^n=(1)/(1024)\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ (1)/(4)/\\\\\left(-(1)/(4)\right)^n=(1)/(4096)\\\\((-1)^n)/(4^n)=(1)/(4^6)\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1`

`(1)/(4^n)=(1)/(4^6)\iff n=6`