Respuesta :

cerverusdante
To fin the [tex] a_{10} [/tex], we need to find the value of r first. And we are going to do that using the formula [tex] a_{n} = a_{1} .r^{n-1} [/tex] 

[tex] a_{4} = a_{1} . r^{4-1} [/tex]
We know that [tex] a_{1} =-1 [/tex] and [tex] a_{4} =64[/tex], so we can replace that in our equation to get:

[tex]64=-1. r^{3} [/tex]
Now we can solve for r:
[tex]r ^{3} = \frac{64}{-1} [/tex]
[tex] r^{3} =-64[/tex]
[tex]r= \sqrt[3]{-64} [/tex]
[tex]r=-4[/tex]

Finally, we can use r to fin [tex] a_{10} [/tex] like this:
[tex] a_{n} = a_{1} . r^{n-1} [/tex]
[tex] a_{10} =(-1) (-4^{10-1} )[/tex]
[tex] a_{10} =(-1)( -4^{9} ) [/tex]
[tex] a_{10} = (-1)(-262144)[/tex]
[tex] a_{10} = 262144[/tex]

We can conclude that the answer is (2) 262,144

Answer:

[tex]a_{10} =262,144[/tex].

Step-by-step explanation:

Given :

In a geometric sequence , [tex]a_{1} = -1\\a_{4} =64[/tex]

To Find : Value of  [tex]a_{10}[/tex]

Solution :

Formula of finding nth term in geometric sequence :

[tex]a_{n} =a_{1} *r^{n-1}[/tex]

Now to find values of r (common ratio)

[tex]a_{4} =a_{1} *r^{4-1}[/tex]

[tex]64 = -1*r^{3}[/tex]

[tex]\sqrt[3]{-64} =r[/tex]

[tex]-4 = r[/tex]

Now Find the value of  [tex]a_{10}[/tex] using formula given above :

[tex]a_{10} =a_{1} *r^{10-1}[/tex]

[tex]a_{10} = -1 *(-4)^{9}[/tex]

[tex]a_{10} = -1 *-262,144[/tex]

[tex]a_{10} =262,144[/tex]

Hence The value of [tex]a_{10} =262,144[/tex].





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