Answer :

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 201;


d1 = a2–a1 = 197–201 = –4


d2 = a3–a2 = 193–197 = –4


The difference in sequence is same and comes to be (–4).


For any term of A.P to be 5


an = a + (n–1)d = 5


an = a + (n–1)d = 201 + (n–1)(–4)


5 = 201–4n + 4


5 = 205–4n


–4n = –205 + 5


n = = 50


The 50th term of A.P is 5


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