Answer :

Formula Used.

d_{n} = a_{n + 1} – a_{n}

a_{n} = a + (n–1)d

In the above sequence,

a = 10;

d_{1} = a_{2}–a_{1} = 15–10 = 5

d_{2} = a_{3}–a_{2} = 20–15 = 5

d_{3} = a_{4}–a_{3} = 25–20 = 5

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

As the last term is 1000

a_{n} = a + (n–1)d

1000 = 10 + (n–1)5

1000 = 10 + 5n–5 = 5 + 5n

5n = 995

n = = 199^{th} term

15 term from the end will be

199 + 1–15 = 185^{th} term

a_{n} = a + (n–1)d

a_{185} = 10 + (185–1)5

= 10 + 184×5

= 10 + 920

= 930

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