Question

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

Answer 1

Hint: First of all the term A.P. here means the arithmetic progression , we also should remember the formula for sum of terms of an arithmetic progression the standard formula for such an addition is given by,
\[{S_n} = \dfrac{{n(n + 1)}}{2}\]
Where $ n $ is the number of the consecutive terms of an arithmetic progression that are added,
The given question asks us to prove that the $ \dfrac{{{S_{3n}}}}{{{S_n}}} = 6 $ holds true for an arithmetic progression in case it follows the rule $ {S_{2n}} = 3{S_n} $
First we will use the rule already given to us to find the number of terms that are actually present in the given arithmetic progression, then we will use that value of $ n $ to prove the given statement.

Complete step-by-step answer:
The Arithmetic progression is a type of series which has a common difference and is formed by adding that difference over and over to the first term. The sum of $ n $ terms of an A.P is given by,
\[{S_n} = \dfrac{{n(n + 1)}}{2}\]
We will now first find the value of $ n $ from the equation,
 $ {S_{2n}} = 3{S_n} $
We proceed as,
\[\dfrac{{2n\left( {2n + 1} \right)}}{2}{\text{ }} = \dfrac{{{\text{ }}3n\left( {n + 1} \right)}}{2}\]
On cancelling $ 2 $ on both sides,
\[2\left( {2n + 1} \right){\text{ }} = {\text{ }}3\left( {n + 1} \right)\]
\[4n + 2{\text{ }} = {\text{ }}3n + 3\]
Which gives us the value of $ n = 1 $
Now we will put this value in the
 $ \dfrac{{{S_{3n}}}}{{{S_n}}} = 6 $
We will solve left hand side and get its value equal to the right hand side,
 $ LHS = \dfrac{{{S_{3n}}}}{{{S_n}}} $
\[\dfrac{{\;\left[ {\dfrac{{3n\left( {3n + 1} \right)}}{2}} \right]}}{{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]}}\]
Put value of $ n = 1 $
We get
 $ LHS = \dfrac{{3 \times 4}}{{1 \times 2}} $
 $ LHS = \dfrac{{12}}{2} $
Which gives
 $ LHS = 6 $
 $ = RHS $
Hence proved.

Note: The sum of an arithmetic progression if the first term and the last term of the arithmetic progression is given is given by
 $ {S_n} = \dfrac{n}{2}(a + l) $
The term $ a $ here means the first term and the $ l $ is the last term of the arithmetic progression, and $ n $ is the number of terms in the progression