Question

If the sum of the first \[2n\] terms of the AP \[2,5,8...\] is equal to the sum of the first \[n\] terms of AP \[57,59,61....\], then \[n\] is equal to
A. \[10\]
B. \[12\]
C. \[11\]
D. \[13\]

Answer 1

Hint: In order to solve the problem, we will be considering both the given series that are in AP. Then we will be applying the formula of the sum of \[n\] terms of an AP to both the series given and obtaining the value. Since we are given that both sum are equal, we will be equating them and upon solving it, we will be obtaining the value of \[n\].

Complete step-by-step solution:
Now let us briefly discuss arithmetic progression. It is a sequence in which the difference between any two consecutive numbers taken would be equal to a constant. We can say that the set of natural numbers follows arithmetic progression as their constant difference is one. The general form of the AP series is \[a,a+d,a+2d,a+3d....a+\left( n-1 \right)d\] where \[d\] is the common difference.
Now let us start solving the problem.
We are given two AP series. They are: \[2,5,8...\] and \[57,59,61....\] but in the first case the number of terms is \[2n\] and in the second case the number of terms is \[n\].
The sum of \[n\] number of terms which are in AP is \[S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
Now let us apply this formula to both of the series.
The sum of the series \[2,5,8...\] with \[2n\] terms is-
\[\begin{align}
  & S=\dfrac{2n}{2}\left( 2\left( 2 \right)+\left( 2n-1 \right)3 \right) \\
 & \Rightarrow S= n\left( 4+6n-3 \right)=n\left( 1+6n \right) \\
\end{align}\]
The sum of the series \[57,59,61....\] with \[n\] number of terms is-
\[\begin{align}
  & S=\dfrac{n}{2}\left( 2\left( 57 \right)+\left( n-1 \right)2 \right) \\
 & \Rightarrow S= \dfrac{n}{2}\left( 114+2n-2 \right)=n\left( 56+n \right) \\
\end{align}\]
Since we are told that the sum of both series are equal, let us equate both sum expressions.
\[\begin{align}
  & n\left( 1+6n \right)=n\left( 56+n \right) \\
 & \Rightarrow 6n-n=55 \\
 & \Rightarrow 5n=55 \\
 & \Rightarrow n=11 \\
\end{align}\]
Therefore, $n=11$. So, option (C) is the correct answer.

Note: Since we were not given the last term of the series nor the number of terms, we have directly opted the formula of sum of series and then found the number of terms each series possesses. While solving for the AP series , we must always check for the common difference of the series as it is the most important factor for solving the series.