Question

Let ${{S}_{n}}$ denote the sum of $n$ terms of an AP. If ${{S}_{2n}}=3{{S}_{n}}$ then the ratio $\dfrac{{{S}_{3n}}}{{{S}_{n}}}=$

1. 4
2. 6
3. 8
4. 10

Answer 1

In this type of question your approach should be greedy, that is whatever the question is saying you just need to do as it mathematically, and you must know some basics of Arithmetic Progression and the sum of terms of Arithmetic Progression formula and you will get your required answer.

Complete step by step answer:
Before solving the given question we must know what an AP is and what its sum formula means.
So, the basic definition of Arithmetic Progression is:
Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
For example, the series of natural numbers: $1,3,5,7,.......$ is an AP, which has a common difference between two successive terms (say $1$ and $3$ ) equal to $2(3-1)$ .
Now since the given question is about the sum of an AP.
So, the sum of $n$ terms of an AP is explained as,
For any progression, the sum of $n$ terms can be easily calculated. For an AP, the sum of the first $n$ terms can be calculated if the first term and the total terms are known. The formula for the arithmetic progression sum is explained below:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)\times d \right]$
Where,
${{S}_{n}}$ denotes the sum of $n$ terms.
$n$ denotes the number of terms.
$a$ denotes the first term.
$d$ denotes the common difference.
So now according to our given question, it is given that
${{S}_{2n}}=3{{S}_{n}}$
That clearly means the sum of $n$ terms of an AP is three times the sum of $2n$ terms of that AP.
So, according to the sum formula of an AP we have,
$\Rightarrow \dfrac{2n}{2}\left[ 2a+(2n-1)d \right]=\dfrac{3n}{2}\left[ 2a+(n-1)d \right]$
Since both sides have $\dfrac{n}{2}$ common therefore it will get eliminated.
So now we have,
$\Rightarrow 2[2a+(2n-1)d]=3[2a+(n-1)d]$
Now on simplifying further,
$\Rightarrow 4a+(4n-2)d=6a+(3n-3)d$
Now after opening brackets and simplify further,
$\Rightarrow 2a=(n+1)d$
So from the given relation in the question we get an equation,
$2a=(n+1)d$ $......(1)$
Now we have to find the ratio given in the question as $\dfrac{{{S}_{3n}}}{{{S}_{n}}}$ ,
So we have,
$\dfrac{{{S}_{3n}}}{{{S}_{n}}}=\dfrac{\dfrac{3n}{2}\left[ 2a+(3n-1)d \right]}{\dfrac{n}{2}\left[ 2a+(n-1)d \right]}$
On cancelling common terms we get,
$\Rightarrow \dfrac{{{S}_{3n}}}{{{S}_{n}}}=\dfrac{3\left[ 2a+(3n-1)d \right]}{\left[ 2a+(n-1)d \right]}$
Now by using equation $(1)$ we get,
$\Rightarrow \dfrac{{{S}_{3n}}}{{{S}_{n}}}=\dfrac{3\left[ (n+1)d+(3n-1)d \right]}{\left[ (n+1)d+(n-1)d \right]}$
Now on simplifying we get,
$\Rightarrow \dfrac{{{S}_{3n}}}{{{S}_{n}}}=\dfrac{3\left[ 4nd \right]}{2nd}$
Now simply cancelling common terms we get,
$\Rightarrow \dfrac{{{S}_{3n}}}{{{S}_{n}}}=6$
Hence we get the required ratio as $\dfrac{{{S}_{3n}}}{{{S}_{n}}}=6$ .

So, the correct answer is “Option 2”.

Note: Arithmetic progression can be applied in real life by analyzing a certain pattern, for example, AP used in straight line depreciation. AP used in prediction of any sequence like when someone is waiting for a cab. Assuming that the traffic is moving at a constant speed he/she can predict when the next cab will come.