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

Solution:

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,

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

$S_{1}=4(1)-(1)^{2}$

$=4-1$

 

$=3$

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

$S_{2}=4(2)-(2)^{2}$

$=8-4$

 

$=4$

Now, as we know,

$a_{n}=S_{n}-S_{n-1}$

So,

$a_{2}=S_{2}-S_{1}$

$=4-3$

$=1$

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

So, for the third term,

$a_{3}=S_{3}-S_{2}$

$=\left[4(3)-(3)^{2}\right]-\left[4(2)-(2)^{2}\right]$

$=(12-9)-(8-4)$

$=3-4$

 

$=-1$

Also, for the tenth term,

$a_{10}=S_{10}-S_{9}$

$=\left[4(10)-(10)^{2}\right]-\left[4(9)-(9)^{2}\right]$

$=(40-100)-(36-81)$

$=-60+45$

 

$=-15$

So, for the nth term,

$a_{n}=S_{n}-S_{n-1}$

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

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

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

 

$=5-2 n$

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