# Q : 11      If the sum of the first $\small n$ terms of an AP is  $\small 4n-n^2$, what is the first term (that is  $\small S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the $\small 3$rd, the $\small 10$th and the $\small n$th terms

It is given that
the sum of the first $\small n$ terms of an AP is  $\small 4n-n^2$
Now,
$\Rightarrow S_n = 4n-n^2$
Now, first term is
$\Rightarrow S_1 = 4(1)-1^2=4-1=3$
Therefore, first term is 3
Similarly,
$\Rightarrow S_2 = 4(2)-2^2=8-4=4$
Therefore, sum of first two terms is 4
Now, we know that
$\Rightarrow S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_2 = \frac{2}{2}\left \{ 2\times 3+(2-1)d \right \}$
$\Rightarrow 4 = \left \{6+d \right \}$
$\Rightarrow d = -2$
Now,
$a_2= a+d = 3+(-2 )= 1$
Similarly,
$a_3= a+2d = 3+2(-2 )= 3-4=-1$
$a_{10}= a+9d = 3+9(-2 )= 3-18=-15$
$a_{n}= a+(n-1)d = 3+(n-1)(-2 )= 5-2n$

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