Question

find the 27th term in the arithmetic sequence of 59,56,53,50

Answer 1

solution:

$A_{n}=59-3(n-1)\\

where, n=1,2,3,.....\\

A1=59\\

A2=56\\

A3=53\\

A_{n}=a+(n-1)d

A27=59+(27-1)3\\

=59-78\\

=-19\\

A27 term is -19$