Question

# The sum of first  $$n$$  terms of an AP is $$3n^2 - n$$. Find the $$25^{th}$$ term of this AP.

Solution

## Given $$S_n=3n^{2}-n$$Then $$S_{n-1}=3\left [ 3(n-1) \right ]^{2}-(n-1)$$$$a_{n}=S_n-S_{n-1}$$$$\Rightarrow a_{n}=\left [ 3n^{2}-n \right ]-\left [ 3(n-1)^{2}-(n-1) \right ]$$$$=3\left [ n^{2}-(n-1)^{2} \right ]-n+n-1$$$$=3(n^{2}-n^{2}+2n-1)-1=3(2n-1)-1=6n-4$$Then, $$a_{25}=6\times 25-4=150-4=146$$Mathematics

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