Question

Let $ {S_n} $ denote the sum of the first ‘n’ terms of an A.P. and $ {S_{2n}} = 3{S_n} $ . Then, the ratio of $ {S_{3n}}:{S_n} $ is equal to
A.4:1
B.6:1
C.8:1
D.10:1

Answer 1

Hint: In the above question, we came across a few terms like Arithmetic Progression that is A.P. and sum of ‘n’ terms denoted by $ {S_n} $ . Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. Common difference is denoted by ‘d’. The finite portion of an AP is known as finite AP and therefore the sum of finite AP is known as arithmetic series. For example: series of even numbers: 2, 4, 6, 8, 10, …... is an AP, which has a common difference between two successive terms that will always be equal. We can find the sum of ‘n’ terms in the series- $ {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] $ .
We will use this formula to solve this question and to find the ratio of $ {S_{3n}}:{S_n} $

Complete step-by-step answer:
 $ {S_n} $ denotes the sum of the first ‘n’ terms of an A.P. So,
 $ {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] $
From the above question, we have $ {S_{2n}} = 3{S_n} $ . So,
 $ {S_{2n}} = \dfrac{{2n}}{2}\left[ {2a + (2n - 1)d} \right] $
According to the question, we will equate the value of $ {S_{2n}} = 3{S_n} $ .
Thus,
     $ {S_{2n}} = 3{S_n} $
 $ \Rightarrow \dfrac{{2n}}{2}\left[ {2a + (2n - 1)d} \right] = 3 \times \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] $
Solving the equation by cancelling out $ \dfrac{n}{2} $ from L.H.S and R.H.S, we get;
 $ \Rightarrow 4a + 2d(2n - 1) = 6a + 3d(n - 1) $
 $ \Rightarrow 4a + 4nd - 2d = 6a + 3nd - 3d $
 $ \Rightarrow 4nd - 2d = 2a + 3nd - 3d $
 $ \therefore 2a = nd + d $
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]}} $
Putting the value of $ 2a = nd + d $ , we get;
  $ \dfrac{{{S_{3n}}}}{{{S_n}}} $ = $ \dfrac{{3\left[ {nd + d + 3nd - d} \right]}}{{nd + d + nd - d}} $
Simplifying the above values, we get;
  $ \dfrac{{{S_{3n}}}}{{{S_n}}} $ = $ \dfrac{{3\left[ {4nd} \right]}}{{2nd}} $
   $ \dfrac{{{S_{3n}}}}{{{S_n}}} $ = 6
 $ {S_{3n}}:{S_n} $ = 6:1
So, the correct answer is “Option B”.

Note: In this question we will put the value of $ {S_n} $ and $ {S_{2n}} $ in $ {S_{2n}} = 3{S_n} $ . We will solve this equation further carefully then we will get the value of 2a. We will further use the value of 2a in getting the ratios of $ {S_{3n}}:{S_n} $ .