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Question: If Sn, the sum of first n terms of an A. P., is given by Sn = \(5{n^2} + 3n\), then find its \({n^{t...

If Sn, the sum of first n terms of an A. P., is given by Sn = 5n2+3n5{n^2} + 3n, then find its nth{n^{th}} term.

Explanation

Solution

To find the nth term an{a_n}, we will use the formula of difference of sums i.e.,
an = Sn – Sn-1 where, Sn is the sum of first n terms of the A. P. and Sn-1 is the sum of the first (n – 1) terms of the A. P.

Complete step-by-step answer:
We are given the sum of first n terms of an A. P. as: Sn = 5n2+3n5{n^2} + 3n
We need to calculate the nth{n^{th}}term of this A. P.
We know the formula of the difference of sums of an A. P. is given by: an=SnSn1{a_n} = {S_n} - {S_{n - 1}} – (1)
Now, for calculating the value of Sn-1, we can say that Sn is true for first n terms then we know that
n > n – 1 , therefore, it will be true for first n – 1 terms as well.
Sn1=5(n1)2+3(n1)\therefore {S_{n - 1}} = 5{\left( {n - 1} \right)^2} + 3\left( {n - 1} \right)
Substituting the values of Sn and Sn-1 in equation (1), we get

\Rightarrow {a_n} = {S_n} - {S_{n - 1}} \\\ \Rightarrow {a_n} = 5{n^2} + 3n - \left\\{ {5{{\left( {n - 1} \right)}^2} + 3\left( {n - 1} \right)} \right\\} \\\

Upon simplification, we get
\Rightarrow {a_n} = 5{n^2} + 3n - \left\\{ {5\left( {{n^2} - 2n + 1} \right) + 3n - 3} \right\\} \\\ \Rightarrow {a_n} = 5{n^2} + 3n - 5{n^2} + 10n - 5 - 3n + 3 \\\ \Rightarrow {a_n} = 10n - 2 \\\
Therefore, the nth term of the A. P., whose sum of first n terms is 5n2+3n5{n^2} + 3n, is found to be 10n – 2.

Note: In such questions, selecting the formula or method to proceed with to calculate the nth{n^{th}} term of the A. P is important. You may go wrong while calculating the sum of first n – 1 terms of the A. P. since it is not provided and you have to show that the sum of the terms will be true for n – 1 terms.