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Let (a­ n ) n ≥ 1 be a sequence such that a­­­ 1 = -1 and 3a n+1 - 3a n =1 for all n ≥ 1. Then find the value of a 2002 -

Let (a­n) n ≥ 1 be a sequence such that a­­­1 = -1 and        3an+1  - 3an =1 for all n ≥ 1. Then find the value of a2002 -

Grade:9

1 Answers

Aditya Gupta
2080 Points
2 years ago
we see that an+1–an =1/3
so it is an AP (arithmetic progression)
now  a2002= a1+2001*1/3
= –1+667
=666

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