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Question: If the sum of first n terms of an A.P. is \(4n-{{n}^{2}}\). What is the first term? Find the sum of ...

If the sum of first n terms of an A.P. is 4n−n24n-{{n}^{2}}. What is the first term? Find the sum of the first 20 terms of an A.P.?

Explanation

Solution

We have given the sum of first n terms of an A.P. which is equal to 4n−n24n-{{n}^{2}}. So, this expression will be Sn{{S}_{n}}. Now, we know that if we have sum of first n terms of an A.P. then we can find the general term of an A.P. by using this relation: Tn=Sn−Sn−1{{T}_{n}}={{S}_{n}}-{{S}_{n-1}}. In this equation, Sn−1{{S}_{n-1}} can be deduced from Sn{{S}_{n}} by substituting n as n – 1 in Sn{{S}_{n}}. Now, the first term can be found by substituting n as 1 in Tn{{T}_{n}}. And the sum of the first 20 terms of an A.P. are found by substituting n as 20 in Sn{{S}_{n}}.

Complete step by step solution:
In the above problem, sum of first n terms of an A.P. are given as:
4n−n24n-{{n}^{2}}
Let us call the above summation as Sn{{S}_{n}}. Now, we can find the general term from this summation expression by using the following relation:
Tn=Sn−Sn−1{{T}_{n}}={{S}_{n}}-{{S}_{n-1}} ……….. (1)
Now, we are going to find Sn−1{{S}_{n-1}} by replacing n as n – 1 in Sn{{S}_{n}} expression and we get,
Sn−1=4(n−1)−(n−1)2 ⇒Sn−1=4n−4−(n2−2n+1) ⇒Sn−1=4n−4−n2+2n−1 ⇒Sn−1=6n−n2−5 \begin{aligned} & {{S}_{n-1}}=4\left( n-1 \right)-{{\left( n-1 \right)}^{2}} \\\ & \Rightarrow {{S}_{n-1}}=4n-4-\left( {{n}^{2}}-2n+1 \right) \\\ & \Rightarrow {{S}_{n-1}}=4n-4-{{n}^{2}}+2n-1 \\\ & \Rightarrow {{S}_{n-1}}=6n-{{n}^{2}}-5 \\\ \end{aligned}
Substituting the above value of Sn−1{{S}_{n-1}} in eq. (1) and we get,
Tn=Sn−Sn−1 ⇒Tn=4n−n2−(6n−n2−5) \begin{aligned} & {{T}_{n}}={{S}_{n}}-{{S}_{n-1}} \\\ & \Rightarrow {{T}_{n}}=4n-{{n}^{2}}-\left( 6n-{{n}^{2}}-5 \right) \\\ \end{aligned}
Opening the brackets after negative sign and we get,
Tn=4n−n2−6n+n2+5{{T}_{n}}=4n-{{n}^{2}}-6n+{{n}^{2}}+5
In the above equation, n2{{n}^{2}} will be cancelled out and we get,
Tn=4n−6n+5 ⇒Tn=−2n+5 \begin{aligned} & {{T}_{n}}=4n-6n+5 \\\ & \Rightarrow {{T}_{n}}=-2n+5 \\\ \end{aligned}
To calculate the first term of the A.P., we are going to put 1 in place of n in the above equation and we get,
⇒T1=−2(1)+5 ⇒T1=−2+5=3 \begin{aligned} & \Rightarrow {{T}_{1}}=-2\left( 1 \right)+5 \\\ & \Rightarrow {{T}_{1}}=-2+5=3 \\\ \end{aligned}
From the above, we got the first terms as 3. Now, to find the sum of first 20 terms we are going to substitute 20 in place of n in 4n−n24n-{{n}^{2}} and we get,
4(20)−(20)2 =80−400 =−320 \begin{aligned} & 4\left( 20 \right)-{{\left( 20 \right)}^{2}} \\\ & =80-400 \\\ & =-320 \\\ \end{aligned}
Hence, we got the sum of the first 20 terms as -320.

Note: To solve the above problem, you should know how to find the general term from the sum of first n terms of an A.P. This is the understanding you should have otherwise you cannot find the first term of the A.P. whereas you can find the sum of first 20 terms of an A.P. using the sum of first n terms expression.