(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]