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If ${{S}_{n}}$ denotes the sum of the first term of an A.P . Prove that ${{S}_{30}}=3({{S}_{20}}-{{S}_{10}})$.

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Hint: Since, ${{S}_{n}}$ denotes the sum of first term of an A.P then ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$. Also, take LHS and put $n=30$and simplify. Then take RHS and put $n=20$and $n=10$ and simplify it. If LHS$=$RHS, you are correct. Try it and you will get the answer.

Complete step-by-step answer:
Arithmetic Progression (A.P) is a sequence of numbers in a particular order. If we observe in our regular lives, we come across progression quite often. For example, roll numbers of a class, days in a week, or months in a year. This pattern of series and sequences has been generalized in maths as progressions. Let us learn here AP definition, important terms such as common difference, the first term of the series, nth term and sum of nth term formulas along with solved questions based on them.
It is a mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as A.P.
The fixed number that must be added to any term of an A.P to get the next term is known as the common difference of the A.P.
An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.
${{n}^{th}}$term of A.P,
${{a}_{n}}=a+(n-1)d$
Now we know for the A.P sum of${{n}^{th}}$the term is,
${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$
where,
$a=$First-term
$d=$ Common difference
$n=$ number of terms
${{a}_{n}}={{n}^{th}}$term
In question we have to prove that ${{S}_{30}}=3({{S}_{20}}-{{S}_{10}})$.
It is given in the question that ${{S}_{n}}$ denotes the sum of the first term of an A.P.
Here, ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$
We have to prove LHS$=$RHS.
So now taking LHS$={{S}_{30}}$
Now since, ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$
putting $n=30$we get,
${{S}_{30}}=\dfrac{30}{2}\left( 2a+(30-1)d \right)=15\left( 2a+29d \right)$
So we get,
LHS$=15\left( 2a+29d \right)$ …………………..(1)
Now take RHS$=3({{S}_{20}}-{{S}_{10}})$.
Putting $n=20$and $n=10$,
we get,
${{S}_{20}}=\dfrac{20}{2}\left( 2a+(20-1)d \right)=10\left( 2a+19d \right)$
${{S}_{10}}=\dfrac{10}{2}\left( 2a+(10-1)d \right)=5\left( 2a+9d \right)$
So now substituting ${{S}_{20}}$ and ${{S}_{10}}$ in RHS we get,
 RHS\[=3(10\left( 2a+19d \right)-5\left( 2a+9d \right))\]
Now simplifying we get,
\[\begin{align}
  & =3\times 5(2\left( 2a+19d \right)-\left( 2a+9d \right)) \\
 & =15(\left( 4a+38d \right)-2a-9d) \\
 & =15(2a+29d) \\
\end{align}\]
RHS\[=15(2a+29d)\]………… (2)
So we get, LHS$=$RHS.
${{S}_{30}}=3({{S}_{20}}-{{S}_{10}})$
Hence proved.

Note: Read the question carefully. Also, you should know that ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$. Don’t make any silly mistakes. You should know the proof. You should also know from where to start. Always keep in mind that signs should not be missed.