# Find the 15th ter

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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