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# If the sum of the first n terms of an A.P. is $4n - {n^2}$, what is the first term (that is ${S_1}$)? What is the sum of the first two terms? What is the second term? Similarly, find the ${3^{rd}}, {10^{th}}$ and the ${n^{th}}$ term.

Last updated date: 06th Sep 2024
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Hint: Use the formula of Arithmetic progression sequence for the nth terms that is ${a_n} = a + \left( {n - 1} \right)d$ where, a initial term of the Arithmetic progression and d is the common difference of successive. Calculate the value of ${S_1}$.

Complete step by step solution:
Given data:
The equation is ${S_n} = 4n - {n^2}$.
Substitute $n = 1$ in ${S_n} = 4n - {n^2}$.
${S_1} = 4\left( 1 \right) - {\left( 1 \right)^2}\\ = 4 - 1\\$
= 3
Hence, the first term of Arithmetic progression is 3. The sum of the first term will be the first term that is ${a_1} = 3$.
Now, substitute $n = 2$ in the expression ${S_n} = 4n - {n^2}$.
${S_2} = 4\left( 2 \right) - {\left( 2 \right)^2}\\ = 8 - 4\\ = 4$
Hence, the sum of the first two terms of Arithmetic progression is 4.
Now, calculate the second term of the Arithmetic progression in the following way.
${S_2} = a + {a_2}\\ 4 = 3 + {a_2}\\ {a_2} = 1$
Hence, the second term is ${a_2} = 1$.

Now, calculate the ${3^{rd}}, {10^{th}}$ and the ${n^{th}}$ term. So, we know about the Arithmetic progression sequence for the ${n^{th}}$ terms is:
${a_n} = {a_1} + \left( {n - 1} \right)d$

Now, calculate the value of ${a_3}$ by substituting the value of ${a_1} = 3,d = - 2\left( {1 - 3} \right),{\rm{ and }}$ n = 3 in ${a_n} = {a_1} + \left( {n - 1} \right)d$.
${a_3} = 3 + \left( {3 - 1} \right)\left( { - 2} \right)\\ = 3 - 4\\ = - 1$
Hence, the third term is ${a_3} = - 1$.
Now, calculate the value of ${a_{10}}$ by substituting the value of ${a_1} = 3,d = - 2\left( {1 - 3} \right),{\rm{ and }}$ n = 10 in ${a_n} = {a_1} + \left( {n - 1} \right)d$.
${a_{10}} = 3 + \left( {10 - 1} \right)\left( { - 2} \right)\\ = 3 - 18\\ = - 15$
Hence, the ${10^{th}}$ term is ${a_{10}} = - 15$
Now, to calculate the value of ${a_n}$, substitute the values ${a_1} = 3,\;d = - 2\left( {1 - 3} \right)\;{\rm{and }}$ n in the expression ${a_n} = {a_1} + \left( {n - 1} \right)d$.
${a_n} = 3 + \left( {n - 1} \right)\left( { - 2} \right)\\ = 3 - 2n + 2\\ = 5 - 2n$

Hence, the nth term is ${a_n} = 5 - 2n$.

Note: Make sure do not use the formula of the sum of n terms in Arithmetic progression that is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where a is initial term of the AP and d is the common difference of successive numbers.