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

1 answer:

5 0

solution:

$A_{n}=59-3(n-1)\\where, n=1,2,3,.....\\A1=59\\A2=56\\A3=53\\A_{n}=a+(n-1)dA27=59+(27-1)3\\=59-78\\=-19\\A27 term is -19$

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