Answer

Verified

450.3k+ 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

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

The 3 + 3 times 3 3 + 3 What is the right answer and class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

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