Question

Given the series 10, 6, 2, -2....find the 13th term?​

Answer 1

$\texttt{10, 6, 2, -2, ...}$

$\texttt{first term, a = 10}$

$\texttt{common difference, d = 6-10 = -4}$

$\mathtt{T_{13} = ?}$

$\mathtt{T_{13} = a+12d = 10+12×(-4) = 10-48}$$\mathtt{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = -38}$