Question.
If the sum of the first $n$ terms of an $A P$ is $4 n-n^{2}$, what is the first term (that is $S_{1}$ ) ? What is the sum of first two terms? What is the second term? Similarly, find the $3 \mathrm{rd}$, the 10 th and the nth terms.
If the sum of the first $n$ terms of an $A P$ is $4 n-n^{2}$, what is the first term (that is $S_{1}$ ) ? What is the sum of first two terms? What is the second term? Similarly, find the $3 \mathrm{rd}$, the 10 th and the nth terms.
Solution:
$\mathrm{S}_{\mathrm{n}}=4 \mathrm{n}-\mathrm{n}^{2}$
Putting $n=1$, we get $S_{1}=4-1=3$
i.e., $\mathrm{t}_{1}=3$
$\mathrm{S}_{2}=4(2)-(2)^{2}=8-4=4$, i.e., $\mathrm{S}_{2}=4$
$\Rightarrow \mathrm{t}_{1}+\mathrm{t}_{2}=4 \Rightarrow 3+\mathrm{t}_{2}=4 \Rightarrow \mathrm{t}_{2}=1$
$t_{2}-t_{1}=1-3=-2 \Rightarrow d=-2$
Then $\mathrm{t}_{3}=\mathrm{t}_{2}+\mathrm{d}=1-2=-1$, i.e., $\mathrm{t}_{3}=-1$
$\mathrm{t}_{10}=\mathrm{a}+9 \mathrm{~d}=3+9(-2)\left(\because \mathrm{t}_{1}=\mathrm{a}\right)$
$\Rightarrow \mathrm{t}_{10}=-15$
$\mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}=3+(\mathrm{n}-1) \times(-2)$
i.e., $\mathrm{t}_{\mathrm{n}}=5-2 \mathrm{n}$
$\mathrm{S}_{\mathrm{n}}=4 \mathrm{n}-\mathrm{n}^{2}$
Putting $n=1$, we get $S_{1}=4-1=3$
i.e., $\mathrm{t}_{1}=3$
$\mathrm{S}_{2}=4(2)-(2)^{2}=8-4=4$, i.e., $\mathrm{S}_{2}=4$
$\Rightarrow \mathrm{t}_{1}+\mathrm{t}_{2}=4 \Rightarrow 3+\mathrm{t}_{2}=4 \Rightarrow \mathrm{t}_{2}=1$
$t_{2}-t_{1}=1-3=-2 \Rightarrow d=-2$
Then $\mathrm{t}_{3}=\mathrm{t}_{2}+\mathrm{d}=1-2=-1$, i.e., $\mathrm{t}_{3}=-1$
$\mathrm{t}_{10}=\mathrm{a}+9 \mathrm{~d}=3+9(-2)\left(\because \mathrm{t}_{1}=\mathrm{a}\right)$
$\Rightarrow \mathrm{t}_{10}=-15$
$\mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}=3+(\mathrm{n}-1) \times(-2)$
i.e., $\mathrm{t}_{\mathrm{n}}=5-2 \mathrm{n}$
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