# The sum of n, 2n and 3n terms of an A.P are ${S_1},{\text{ }}{S_2},{\text{ }}{S_3}$ respectively.

Prove that ${S_3}{\text{ }} = {\text{ }}3({S_2}{\text{ }} - {\text{ }}{S_1})$.

Last updated date: 17th Mar 2023

•

Total views: 306k

•

Views today: 3.86k

Answer

Verified

306k+ views

Hint: First find the value of \[{S_1},{\text{ }}{S_2}\] and \[{S_3}\]. And then subtract \[{S_1}\] from \[{S_2}\]. After that multiply the equation by 3 to prove the result.

Let a and d be the first term and common difference of the given A.P.

As we know that sum of t terms of this A.P will be,

Let, \[{{\text{S}}_t}\] be the sum of the first t terms of A.P. then,

\[ \Rightarrow {S_t}{\text{ }} = {\text{ }}\dfrac{t}{2}[2a{\text{ }} + {\text{ }}(t{\text{ }} - {\text{ }}1)d]\]

As, we know that \[{{\text{S}}_1}\] is the sum of n terms of A.P. So,

\[ \Rightarrow {S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(n{\text{ }} - {\text{ }}1)d]\] (1)

As, we know that \[{{\text{S}}_2}\] is the sum of 2n terms of A.P. So,

\[ \Rightarrow {S_2}{\text{ }} = {\text{ }}\dfrac{{2n}}{2}[2a{\text{ }} + {\text{ }}(2n{\text{ }} - {\text{ }}1)d]\] (2)

As, we know that \[{{\text{S}}_3}\] is the sum of 3n terms of A.P. So,

\[ \Rightarrow {S_3}{\text{ }} = {\text{ }}\dfrac{{3n}}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]\] (3)

Now, we know that we had to prove \[{S_3}{\text{ }} = {\text{ }}3({S_2}{\text{ }} - {\text{ }}{S_1})\].

So, subtracting equation 1 from 2. We get,

\[ \Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}n[2a{\text{ }} + {\text{ }}(2n{\text{ }} - {\text{ }}1)d]{\text{ }} - {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(n{\text{ }} - {\text{ }}1)d]{\text{ }}\]

On, solving above equation becomes,

\[ \Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}n\left\{ {\dfrac{{4a{\text{ }} + {\text{ }}4nd{\text{ }} - {\text{ }}2d{\text{ }} - {\text{ }}2a{\text{ }} - {\text{ }}nd{\text{ }} + {\text{ }}d}}{2}} \right\}\]

On, solving above equation becomes,

\[ \Rightarrow {S_2} - {\text{ }}{S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}\left[ {2a{\text{ }} + {\text{ }}3nd{\text{ }} - {\text{ }}d} \right]\]

Now, reducing the above equation to prove the result. We get,

\[ \Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]\]

Multiplying both sides of the above equation by 3. We get,

\[ \Rightarrow 3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}\dfrac{{3n}}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]\]

Now, using equation 3. We can write above equation as,

\[ \Rightarrow 3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}{S_3}\]

Hence, \[3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}{S_3}\]

Note: Whenever we came up with this type of question then easiest and efficient way is to first, assume first term and common difference of the A.P. and then find sum of n, 2n and 3n terms of the A.P, using formula of sum of t terms of A.P and then put their values to get required result.

Let a and d be the first term and common difference of the given A.P.

As we know that sum of t terms of this A.P will be,

Let, \[{{\text{S}}_t}\] be the sum of the first t terms of A.P. then,

\[ \Rightarrow {S_t}{\text{ }} = {\text{ }}\dfrac{t}{2}[2a{\text{ }} + {\text{ }}(t{\text{ }} - {\text{ }}1)d]\]

As, we know that \[{{\text{S}}_1}\] is the sum of n terms of A.P. So,

\[ \Rightarrow {S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(n{\text{ }} - {\text{ }}1)d]\] (1)

As, we know that \[{{\text{S}}_2}\] is the sum of 2n terms of A.P. So,

\[ \Rightarrow {S_2}{\text{ }} = {\text{ }}\dfrac{{2n}}{2}[2a{\text{ }} + {\text{ }}(2n{\text{ }} - {\text{ }}1)d]\] (2)

As, we know that \[{{\text{S}}_3}\] is the sum of 3n terms of A.P. So,

\[ \Rightarrow {S_3}{\text{ }} = {\text{ }}\dfrac{{3n}}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]\] (3)

Now, we know that we had to prove \[{S_3}{\text{ }} = {\text{ }}3({S_2}{\text{ }} - {\text{ }}{S_1})\].

So, subtracting equation 1 from 2. We get,

\[ \Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}n[2a{\text{ }} + {\text{ }}(2n{\text{ }} - {\text{ }}1)d]{\text{ }} - {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(n{\text{ }} - {\text{ }}1)d]{\text{ }}\]

On, solving above equation becomes,

\[ \Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}n\left\{ {\dfrac{{4a{\text{ }} + {\text{ }}4nd{\text{ }} - {\text{ }}2d{\text{ }} - {\text{ }}2a{\text{ }} - {\text{ }}nd{\text{ }} + {\text{ }}d}}{2}} \right\}\]

On, solving above equation becomes,

\[ \Rightarrow {S_2} - {\text{ }}{S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}\left[ {2a{\text{ }} + {\text{ }}3nd{\text{ }} - {\text{ }}d} \right]\]

Now, reducing the above equation to prove the result. We get,

\[ \Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]\]

Multiplying both sides of the above equation by 3. We get,

\[ \Rightarrow 3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}\dfrac{{3n}}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]\]

Now, using equation 3. We can write above equation as,

\[ \Rightarrow 3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}{S_3}\]

Hence, \[3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}{S_3}\]

Note: Whenever we came up with this type of question then easiest and efficient way is to first, assume first term and common difference of the A.P. and then find sum of n, 2n and 3n terms of the A.P, using formula of sum of t terms of A.P and then put their values to get required result.

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