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 3rd, the \small 10th and the \small nth terms

Answers (1)

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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