What is the 21st term in the following geometric sequence? –8, 16, –32, 64, –128, . . Jul 17 | Aber from Burlington, IA | 2 Answers | 0 Votes Mark favorite Subscribe Comment
-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 Jul 17 | Francisco E. Comment
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. Jul 17 | Philip P. Comment