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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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