KVPY Sample Paper KVPY Stream-SX Model Paper-15

Let \[\{{{a}_{n}}\}\,\,(n\ge 1)\] be a sequence such that \[{{a}_{1}}=1,\] and \[3{{a}_{n+1}}-3{{a}_{n}}=1\] for all \[n\ge 1.\] Then \[{{a}_{2002}}\] is equal to

A) 666

B) 667

C) 668

D) 669

Correct Answer: C

Solution :

\[{{a}_{1}}=1\]

\[3{{a}_{n+1}}-3{{a}_{n}}=1\]

\[{{a}_{n+1}}=\frac{3{{a}_{n}}+1}{3}={{a}_{n}}+\frac{1}{3},\]

\[{{a}_{2}}={{a}_{1}}+\frac{1}{3}=1+\frac{1}{3}\]

\[{{a}_{3}}={{a}_{2}}+\frac{1}{3}={{a}_{1}}+\frac{1}{3}+\frac{1}{3}=1+\frac{2}{3}\]

\[{{a}_{4}}={{a}_{3}}+\frac{1}{3}=1+\frac{2}{3}+\frac{1}{3}=1+\frac{3}{3}\]

?. ??? ???? ??..

\[{{a}_{2002}}=1+\frac{2001}{3}=1+667=668\]