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Question: 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...

If the sum of the first n terms of an A.P. is 4n−n24n - {n^2}, what is the first term (that is S1{S_1})? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd,10th{3^{rd}}, {10^{th}} and the nth{n^{th}} term.

Explanation

Solution

Use the formula of Arithmetic progression sequence for the nth terms that is an=a+(n−1)d{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 S1{S_1}.

Complete step by step solution:
Given data:
The equation is Sn=4n−n2{S_n} = 4n - {n^2}.
Substitute n=1n = 1 in Sn=4n−n2{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 a1=3{a_1} = 3.
Now, substitute n=2n = 2 in the expression Sn=4n−n2{S_n} = 4n - {n^2}.
S2=4(2)−(2)2 =8−4 =4{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.
S2=a+a2 4=3+a2 a2=1{S_2} = a + {a_2}\\\ 4 = 3 + {a_2}\\\ {a_2} = 1
Hence, the second term is a2=1{a_2} = 1.

Now, calculate the 3rd,10th{3^{rd}}, {10^{th}} and the nth{n^{th}} term. So, we know about the Arithmetic progression sequence for the nth{n^{th}} terms is:
an=a1+(n−1)d{a_n} = {a_1} + \left( {n - 1} \right)d

Now, calculate the value of a3{a_3} by substituting the value of a1=3,d=−2(1−3),and{a_1} = 3,d = - 2\left( {1 - 3} \right),{\rm{ and }} n = 3 in an=a1+(n−1)d{a_n} = {a_1} + \left( {n - 1} \right)d.
a3=3+(3−1)(−2) =3−4 =−1{a_3} = 3 + \left( {3 - 1} \right)\left( { - 2} \right)\\\ = 3 - 4\\\ = - 1
Hence, the third term is a3=−1{a_3} = - 1.
Now, calculate the value of a10{a_{10}} by substituting the value of a1=3,d=−2(1−3),and{a_1} = 3,d = - 2\left( {1 - 3} \right),{\rm{ and }} n = 10 in an=a1+(n−1)d{a_n} = {a_1} + \left( {n - 1} \right)d.
a10=3+(10−1)(−2) =3−18 =−15{a_{10}} = 3 + \left( {10 - 1} \right)\left( { - 2} \right)\\\ = 3 - 18\\\ = - 15
Hence, the 10th{10^{th}} term is a10=−15{a_{10}} = - 15
Now, to calculate the value of an{a_n}, substitute the values a1=3,  d=−2(1−3)  and{a_1} = 3,\;d = - 2\left( {1 - 3} \right)\;{\rm{and }} n in the expression an=a1+(n−1)d{a_n} = {a_1} + \left( {n - 1} \right)d.
an=3+(n−1)(−2) =3−2n+2 =5−2n{a_n} = 3 + \left( {n - 1} \right)\left( { - 2} \right)\\\ = 3 - 2n + 2\\\ = 5 - 2n

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

Note: Make sure do not use the formula of the sum of n terms in Arithmetic progression that is Sn=n2[2a+(n−1)d]{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.