If the sum of the first $^{'} n^{'}$ terms of an $A P$ is $4 n - n^{2}$ . What is the sum of first two terms? Find the $3^{r d}$ and the $10^{t h}$ terms.

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Solution

$\begin{matrix} S_{n} = 4 n - n^{2} p u t, n = 1 S_{1} = 4 (1) - {(1)}^{2} S_{1} = 4 - 1 = 3 t h e n, a = 3 n = 2 S_{2} = 4 (2) - {(2)}^{2} S_{2} = 8 - 4 = 4 S_{2} = 4 f i r s t + sec o n d = 4 3 + a_{2} = 4 a_{2} = 4 - 3 = 1 f i n d c o m m o n d i f f e r e n c e (d) = a_{2} - a a_{n} = a + (n - 1) d 3 r d t e r m = a_{3} = a + 2 d = 3 + 2 (- 2) = 3 - 4 = - 1 10 t h t e r m a_{10} = a + 9 d = 3 - + 9 (\begin{matrix} - 2 = 3 - 18 \end{matrix}) = - 15 \end{matrix}$

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If the sum of the first ′n′ terms of an AP is 4n−n2. What is the sum of first two terms? Find the 3rd and the 10th terms.

Sn=4n−n2put,n=1S1=4(1)−(1)2S1=4−1=3then,a=3n=2S2=4(2)−(2)2S2=8−4=4S2=4first+second=43+a2=4a2=4−3=1findcommondifference(d)=a2−aan=a+(n−1)d3rdterm=a3=a+2d=3+2(−2)=3−4=−110thterma10=a+9d=3−+9(−2=3−18)=−15

If the sum of the first $^{'} n^{'}$ terms of an $A P$ is $4 n - n^{2}$ . What is the sum of first two terms? Find the $3^{r d}$ and the $10^{t h}$ terms.