Answer :

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 8;


d1 = a2–a1 = 11–8 = 3


d2 = a3–a2 = 14–11 = 3


d3 = a4–a3 = 17–14 = 3


The difference in sequence is same and comes to be (3).


For any term of A.P to be 272


an = a + (n–1)d = 272


an = a + (n–1)d = 8 + (n–1)(3)


= 8 + 3n–3


= 5 + 3n


5 + 3n = 272


3n = 272–5 = 267


n = = 89


The 89th term of A.P has value 272


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