If the sum of the first n terms of an A.P is 4n − n2,


If the sum of the first $n$ terms of an A.P is $4 n-n^{2}$, What is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the $n$th terms.


In the given problem, the sum of n terms of an A.P. is given by the expression,

$S_{n}=4 n-n^{2}$

So here, we can find the first term by substituting,

$S_{n}=4 n-n^{2}$





Similarly, the sum of first two terms can be given by,





Now, as we know,






Now, using the same method we have to find the third, tenth and nth term of the A.P.

So, for the third term,







Also, for the tenth term,







So, for the nth term,



$=\left(4 n-n^{2}\right)-\left(4 n-4-n^{2}-1+2 n\right)$

$=4 n-n^{2}-4 n+4+n^{2}+1-2 n$


$=5-2 n$

Therefore, $a=3, S_{2}=4, a_{2}=1, a_{3}=-1, a_{10}=-15$.

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