MATH SOLVE

5 months ago

Q:
# (9.) In a geometric sequence, it is known that a1= -1 and a4= 64. The value of a10 is...

Accepted Solution

A:

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

[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