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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 -

Antareep Dey , 7 Years ago
Grade 9
anser 1 Answers
Aditya Gupta
we see that an+1–an =1/3
so it is an AP (arithmetic progression)
now  a2002= a1+2001*1/3
= –1+667
=666
Last Activity: 7 Years ago
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