If ${{S}_{n}}$ denotes the sum of the first term of an A.P . Prove that ${{S}_{30}}=3({{S}_{20}}-{{S}_{10}})$.
Last updated date: 24th Mar 2023
•
Total views: 307.8k
•
Views today: 8.85k
Answer
307.8k+ views
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.
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.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
