
The sum of n, 2n and 3n terms of an A.P are ${S_1},{\text{ }}{S_2},{\text{ }}{S_3}$ respectively.
Prove that ${S_3}{\text{ }} = {\text{ }}3({S_2}{\text{ }} - {\text{ }}{S_1})$.
Answer
515.1k+ 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
What percentage of the area in India is covered by class 10 social science CBSE

The area of a 6m wide road outside a garden in all class 10 maths CBSE

What is the electric flux through a cube of side 1 class 10 physics CBSE

If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE

The radius and height of a cylinder are in the ratio class 10 maths CBSE

An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE

Trending doubts
What did the military generals do How did their attitude class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

For Frost what do fire and ice stand for Here are some class 10 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

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
