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Question

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


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

Mathematics

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