Question

If Sn equal to 3n square minus n square is the sum of nth term of an AP. determine the AP and its 24th term​

Answer 1

$\rm{a_{n}=6n-4,\ a_{24}=140}$

$\Large\textrm{Proper question.}$

New equation

$\rm{\bigstar S_{n}=3n^{2}-n^{2}\longrightarrow S_{n}=3n^{2}-n}$

If this is not the required answer, please view the passage. You will surely find the answer from the explanation.

$\Large\textrm{Formulae.}$

The formula of,

$\rm{1.\longrightarrow\boxed{\rm{S_{n}-S_{n-1}=a_{n}\ (n\geq2)}}}$

The formula of,

$\rm{2.\longrightarrow\boxed{\rm{S_{1}=a_{1}}}}$

$\Large\textrm{Formula 1}$

The sum of the first single term includes only one term.

$\Large\textrm{Formula 2}$

The formula 2 arises from the definition of the sum.

$\rm{S_{n}}$ denotes the sum of the first $\rm{n}$ terms. So we would get the following equations.

$\rm{S_{n}=a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}\ \cdots\ [1]}$

$\rm{S_{n-1}=a_{1}+a_{2}+\cdots+a_{n-1}\ \cdots\ [2]}$

To eliminate the common terms, we subtract the equations 1 and 2, The result we get is the formula presented above.

$\rm{\therefore S_{n}-S_{n-1}=a_{n}}$

$\Large\textrm{Explained.}$

$\rm{S_{n}=3n^{2}-n}$

$\therefore a_{1}=2$

$\rm{S_{n}=3n^{2}-n\ \cdots\ [1]}$

$\rm{S_{n-1}=3(n-1)^{2}-(n-1)}$

$\rm{S_{n-1}=(3n^{2}-6n+3)-(n-1)}$

$\rm{S_{n-1}=3n^{2}-7n+4\ \cdots\ [2]}$

$[1]-[2]$

$\rm{\therefore a_{n}=6n-4\ (n\geq 2)}$

Hence shown that

$\rm{a_{n}=6n-4}$

$\rm{(n=24)}$

$a_{24}=144-4=140$