Question

If the sum of the first n terms of an A.P. is \[4n - {n^2}\], what is the first term (that is \[{S_1}\])? 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.

Answer 1

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.