–8, 16, –32, 64, –128, . .

^{n-1}), being a = 8 and r ( common ratio) = -2

^{(21-1)}= 8 x (-2)

^{20}= -8388608

What is the 21st term in the following geometric sequence?

–8, 16, –32, 64, –128, . .

–8, 16, –32, 64, –128, . .

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-8, 16, –32, 64, –128

This corresponds to the following form of sequence:

a X (-2^{n-1}), being a = 8 and r ( common ratio) = -2

then the 21 term will be

a X r^{(21-1)}= 8 x (-2)^{20} = -8388608

The terms in the sequence are:

-2^{n-1}*8, n=1,2,3,... (Since 8=2^{3},we could also write the sequence as -2^{n+2}, n=1,2,3,...)

The 21st term occurs for n=21:

-2^{21-1}*8 = -2^{20}*8

Use your calculator to compute the answer.

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