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If the arithmetic progression whose common difference is non zero, the sum of first 3n terms is equal to the sum of next n terms. Then, the ratio of the sum of the first 2n terms to next 2n terms is
\[\begin{align}
  & A.\dfrac{1}{5} \\
 & B.\dfrac{2}{5} \\
 & C.\dfrac{3}{4} \\
 & D.\text{ None of these} \\
\end{align}\]

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Last updated date: 17th Jul 2024
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Answer
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Hint: The ${{n}^{th}}$ term of a sequence having common difference d and first term a is
\[{{a}_{n}}=a+\left( n-1 \right)d\]
Where n is the total number of terms.
In our question, we have \[\text{3n}+\text{n}=\text{4n}\] terms in the sequence. It is given that, sum of first 3n and next n terms is the same, so we will use the formula below. Also, the sum of n terms of an AP (arithmetic progression) is given by: \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
Where, a is the first term and n is the number of terms and d is a common difference.

Complete step-by-step answer:
Given that, the sum of 3n terms and n terms are used in the question.
There are a total of \[\text{3n}+\text{n}=\text{4n}\] terms in the sequence.
The ${{n}^{th}}$ term of a sequence having common difference d and first term a is
\[{{a}_{n}}=a+\left( n-1 \right)d\]
Where n is the total number of terms.
Also, the sum of n terms of an AP (arithmetic progression) is given by
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
Where, a is the first term and n is the number of terms and d is a common difference.
We are given that, sum of the first 3n terms is equal to the sum of the next n terms.
Sum of 3n terms of AP = ${{S}_{3n}}$ = using above formula, we get:
\[{{S}_{3n}}=\dfrac{3n}{2}\left[ 2a+\left( 3n-1 \right)d \right]\]
Next n term is given by \[\text{3n}+\text{n}=\text{4n}\]
Then, according to condition of question, we have:
\[\begin{align}
  & {{S}_{3n}}={{S}_{4n}}-{{S}_{3n}} \\
 & \Rightarrow 2{{S}_{3n}}={{S}_{4n}} \\
\end{align}\]
Substituting the values of sum given in formula above, we get:
\[2\left( \dfrac{3n}{2}\left[ 2a+\left( 3n-1 \right)d \right] \right)=\left( \dfrac{4n}{2}\left[ 2a+\left( 4n-1 \right)d \right] \right)\]
Cancelling 2 and solving further, we get:
\[3n\left[ 2a+\left( 3n-1 \right)d \right]=2n\left[ 2a+\left( 4n-1 \right)d \right]\]
Cancelling n we get:
\[\begin{align}
  & 3n\left[ 2a+\left( 3n-1 \right)d \right]=2n\left[ 2a+\left( 4n-1 \right)d \right] \\
 & \Rightarrow 6a+9nd-3d=4a+8nd-2d \\
 & \Rightarrow 6a-4a=3d-2d+8nd-9nd \\
 & \Rightarrow 2a=d+\left( -nd \right) \\
 & \Rightarrow 2a=d\left( 1-n \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
Finally we have to find the ratio of sum of first 2n terms to next 2n terms;
Next 2n term is given by,
\[{{S}_{2n}}={{S}_{4n}}-{{S}_{2n}}\]
Then, the ratio is \[\dfrac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}\]
Substituting the value of formula of sum of n terms stated above, we get:
\[\dfrac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}=\dfrac{\dfrac{2n}{2}\left[ 2a+\left( 2n-1 \right)d \right]}{\dfrac{4n}{2}\left[ 2a+\left( 4n-1 \right)d \right]-\dfrac{2n}{2}\left[ 2a+\left( 2n-1 \right)d \right]}\]
Cancelling the common terms, we get:
\[\begin{align}
  & \dfrac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}=\dfrac{\left[ 2a+\left( 2n-1 \right)d \right]}{2\left[ 2a+\left( 4n-1 \right)d \right]-\left[ 2a+\left( 2n-1 \right)d \right]} \\
 & \Rightarrow \dfrac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}=\dfrac{\left[ 2a+\left( 2n-1 \right)d \right]}{2\left[ 20+\left( 4n-1 \right)d \right]-\left[ 2a+\left( 2n-1 \right)d \right]} \\
\end{align}\]
Using equation (i) and substituting 2a=(1-n)d we get,
\[\begin{align}
  & \dfrac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}=\dfrac{\left( 1-n \right)d+\left( 2n-1 \right)d}{2\left( \left( 1-n \right)d+\left( 4n-1 \right)d \right)-\left( 1-n \right)d+\left( 2n-1 \right)d} \\
 & \Rightarrow \dfrac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}=\dfrac{nd}{2\left( nd \right)-nd} \\
 & \Rightarrow \text{Ratio }\dfrac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}=\dfrac{nd}{5nd} \\
\end{align}\]
Ratio of sum of first 2n terms to next 2n terms is $\dfrac{1}{5}$

So, the correct answer is “Option A”.

Note: The biggest possibility of mistake in this question is at the point where it is given in the question that the sum of first 2n terms to next 2n terms is to be taken. The term ' next 2n term' does not really mean to take ${{S}_{2n}}$ it means that the original 4n terms and next 2n term difference is to be taken. That is why, we took ${{S}_{4n}}-{{S}_{2n}}$ for the next 2n term. This is the very important key point in the question.