we can write this in terms of regular expressions

like a^{1}b^{+}c^{1} + a^{2}b^{+}c^{2} +.............+ a^{99}b^{+}c^{99 }, this is finite, So it should be regular

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If there would have been no upper bound on $n$, then $L$ would be context free. Here there is an upper bound to $n$ and therefore our language here is finite. All finite languages are regular. Also, all regular languages are context free.

Therefore given language is regular and context free. Ans is (C)

Therefore given language is regular and context free. Ans is (C)

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