If the sum of n terms of a sequence is S_n = | vedclass

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(A) Given the sum of $n$ terms $S_n = 3n^2 + 4n$. The $n^{th}$ term $a_n$ is given by $a_n = S_n - S_{n-1}$. $a_n = (3n^2 + 4n) - (3(n-1)^2 + 4(n-1))$

Vedclass · Accepted Answer

Arithmetic Progression

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(A) Given the sum of $n$ terms $S_n = 3n^2 + 4n$. The $n^{th}$ term $a_n$ is given by $a_n = S_n - S_{n-1}$. $a_n = (3n^2 + 4n) - (3(n-1)^2 + 4(n-1))$. $a_n = (3n^2 + 4n) - (3(n^2 - 2n + 1) + 4n - 4)$. $a_n = 3n^2 + 4n - (3n^2 - 6n + 3 + 4n - 4)$. $a_n = 3n^2 + 4n - (3n^2 - 2n - 1)$. $a_n = 6n + 1$. Since the $n^{th}$ term $a_n = 6n + 1$ is a linear expression in $n$,the common difference $d = a_n - a_{n-1} = (6n + 1) - (6(n-1) + 1) = 6$,which is a constant. Therefore,the sequence is an Arithmetic Progression.

If the sum of $n$ terms of a sequence is $S_n = 3n^2 + 4n$,then what type of sequence is it?

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