Answer :

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 10;


d1 = a2–a1 = 15–10 = 5


d2 = a3–a2 = 20–15 = 5


d3 = a4–a3 = 25–20 = 5


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


As the last term is 1000


an = a + (n–1)d


1000 = 10 + (n–1)5


1000 = 10 + 5n–5 = 5 + 5n


5n = 995


n = = 199th term


15 term from the end will be


199 + 1–15 = 185th term


an = a + (n–1)d


a185 = 10 + (185–1)5


= 10 + 184×5


= 10 + 920


= 930


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