The 100th term of 1, 7, 13, 19, … is

asked May 14, 2023 160k views

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$Consider \: the \: sequence \: u_(n) \\ u_(n) \: appears \: to \: be \: arithmtic$

$u_(4)-u_(3)=u_(3)-u_(2)=... \\ 19 - 13 = 13 - 7 = .. \\ 6 = 6 = ...$

$u_(n) \: is \: arithmetic \: with \: d = 6 \:and \: u_(1) = 1 \\$

$u_(n) = u_(1)+(n-1)d \\ u_(100) = 1 + (100 - 1)6 \\ u_(100) = 595$

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