Answer :

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 17;


d1 = a2–a1 = 22–17 = 5


d2 = a3–a2 = 27–22 = 5


d3 = a4–a3 = 32–27 = 5


As in A.P the difference between the 2 terms is always constant


The difference in sequence is same and comes to be 5.


The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 17 + (n–1)5


= 17 + 5n–5


= 12 + 5n


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