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Let {an}(ngeq1) be a sequence such that ...

Let `{a_n}(ngeq1)` be a sequence such that `a_1=1,a n d3a_(n+1)-3a_n=1fora l lngeq1.` Then find the value of `a_(2002.)`

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The correct Answer is:
668
`a_(1)=1`
`3a_(n+1)-3a_(n)=1`
`thereforea_(n+1)=(2a_(n)+1)/3=a_(n)+1/3`
`thereforea_(2)=a_(1)+1/3=1+1/3`
`thereforea_(3)=a_(2)+1/3=a_(1)+1/3+1/3=1+2/3`
`thereforea_(4)=a_(3)+1/3=1+2/3+1/3=1+3/3`
…. ….. …..
…. ….. ……
`thereforea_(2002)=1+2001/3=1+667=668`
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