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Grade: 9
        
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 -
7 months ago

Answers : (1)

Aditya Gupta
1154 Points
							
we see that an+1–an =1/3
so it is an AP (arithmetic progression)
now  a2002= a1+2001*1/3
= –1+667
=666
7 months ago
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