Question

Let \[{S_n}\] denote the sum of first \[n\] terms of an A.P. If \[{S_{2n}} = 3{S_n}\] , then prove that \[\dfrac{{{S_{3n}}}}{{{S_n}}} = 6\] .

Answer 1

Hint: Here we will first find the sum \[2n\] terms of an AP using the formula of sum of \[n\] terms of an AP. Then we will substitute the values in the given condition and simplify it further to get the value of first terms of the AP in terms of common difference. We will substitute this value in the equation that needs to be found out. Finally, we will solve the equation and get the required answer.

Formula Used:
Arithmetic Sum Formula: \[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where, \[n = \] number of terms, \[a = \] First term and \[d = \] Common difference.

Complete step by step solution:
We have to prove:
\[\dfrac{{{S_{3n}}}}{{{S_n}}} = 6\]……\[\left( 1 \right)\]
Let the first term be \[a\] and the common difference is \[d\] of the A.P.
We know the formula for sum of \[n\] terms is:
\[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\]……\[\left( 2 \right)\]
Similarly for getting sum of \[2n\] terms we will replace \[n\] by \[2n\] in above equation and get,
 \[{S_n} = \dfrac{{2n}}{2}\left( {2a + \left( {2n - 1} \right)d} \right)\]…..\[\left( 3 \right)\]
It is given that \[{S_{2n}} = 3{S_n}\].
So, substituting value from equation \[\left( 2 \right)\] and \[\left( 3 \right)\] in above condition, we get
\[ \Rightarrow \dfrac{{2n}}{2}\left( {2a + \left( {2n - 1} \right)d} \right) = \dfrac{{3n}}{2}\left( {2a + \left( {n - 1} \right)d} \right)\]
\[ \Rightarrow 2n\left( {2a + \left( {2n - 1} \right)d} \right) = 3n\left( {2a + \left( {n - 1} \right)d} \right)\]
Cancelling \[n\] from both sides, we get
 \[ \Rightarrow 2\left( {2a + \left( {2n - 1} \right)d} \right) = 3\left( {2a + \left( {n - 1} \right)d} \right)\]
Multiplying the terms, we get
\[ \Rightarrow 4a + 2d\left( {2n - 1} \right) = 6a + 3d\left( {n - 1} \right)\]
Rewriting the above equation, we get
\[ \Rightarrow 6a - 4a = 2d\left( {2n - 1} \right) - 3d\left( {n - 1} \right)\]
Multiplying the terms, we get
\[\begin{array}{l} \Rightarrow 2a = 4dn - 2d - 3dn + 3d\\ \Rightarrow 2a = dn + d\end{array}\]
Taking \[d\] common in right hand side, we get
\[ \Rightarrow 2a = d\left( {n + 1} \right)\]…..\[\left( 4 \right)\]
Now, using the \[{n^{th}}\] term formula on left hand side of equation \[\left( 1 \right)\], we get
\[\begin{array}{l}\dfrac{{{S_{3n}}}}{{{S_n}}} = \dfrac{{\dfrac{{3n}}{2}\left( {2a + \left( {3n - 1} \right)d} \right)}}{{\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)}}\\ \Rightarrow \dfrac{{{S_{3n}}}}{{{S_n}}} = \dfrac{{3n\left( {2a + \left( {3n - 1} \right)d} \right)}}{{n\left( {2a + \left( {n - 1} \right)d} \right)}}\end{array}\]
Now substituting the value from equation \[\left( 4 \right)\] in above equation, we get
\[ \Rightarrow \dfrac{{{S_{3n}}}}{{{S_n}}} = \dfrac{{3n\left( {d\left( {n + 1} \right) + d\left( {3n - 1} \right)} \right)}}{{n\left( {d\left( {n + 1} \right) + d\left( {n - 1} \right)} \right)}}\]
Cancelling \[n\] from both numerator and denominator, we get
\[ \Rightarrow \dfrac{{{S_{3n}}}}{{{S_n}}} = \dfrac{{3\left( {d\left( {n + 1} \right) + d\left( {3n - 1} \right)} \right)}}{{\left( {d\left( {n + 1} \right) + d\left( {n - 1} \right)} \right)}}\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{{{S_{3n}}}}{{{S_n}}} = \dfrac{{3\left( {dn + d + 3nd - d} \right)}}{{\left( {dn + d + dn - d} \right)}}\]
Adding and subtracting the terms, we get
\[ \Rightarrow \dfrac{{{S_{3n}}}}{{{S_n}}} = \dfrac{{3\left( {4nd} \right)}}{{\left( {2dn} \right)}}\]
Cancelling the similar terms, we get
\[\begin{array}{l} \Rightarrow \dfrac{{{S_{3n}}}}{{{S_n}}} = 3 \times 2\\ \Rightarrow \dfrac{{{S_{3n}}}}{{{S_n}}} = 6\end{array}\]
So, we get the Right Hand side by solving the Left hand side.
Hence Proved

Note:
Progression in general means a series or sequence in which a number is arranged in such a way that there is a relation between the consecutive terms. Arithmetic progression is the sequence of terms where the common difference between the successive members is constant. If we have to find the sum of the A.P it is known as Arithmetic Series. There are various real life examples of progression such as, Roll numbers of the students, days in a week or in a month.