Question

# If $$t_{n}$$ denotes the $$n$$th term of the series $$2+3+6+11+18+...$$then $$t_{50}$$ is

A
4921
B
492
C
502+1
D
492+2

Solution

## The correct option is C $$49^{2}+2$$In the given series, the difference between the successive terms are in A.P.Therefore, $$t_n$$ will be of the form $$an^2+bn+c$$Now,        $$t_1=2=a+b+c$$        $$t_2=3=4a+2b+c$$        $$t_3=6=9a+3b+c$$Solving for $$a$$, $$b$$ and $$c$$We get  $$a=1$$, $$b=-2$$ and $$c=3$$Therefore, $$t_n=n^2-2n+3=(n-1)^2+2$$Thus, $$t_{50} =49^2+2$$Ans: DMathematics

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