Question

A set of $56$ tuning forks are so arranged in series that each fork gives $4$ beats per second with the previous one. The frequency of the last fork is $3$ times that of the first. The frequency of the first fork is
A.$110$
B.$60$
C.$56$
D.$52$

Answer 1

Hint: If we observe carefully then the question says that the beats per second of the tuning forks are such that they form an arithmetic progression because, with a common difference of $4$ between all the terms. The formula for $n\text{th}$ of the arithmetic progression can be used to solve this question.

Complete answer:
Let us assume that the frequency of the first fork is $f$, then using our knowledge of arithmetic progression we can write down the frequency of the last fork as follows:
$\begin{align}
  & f'=f+\left( 56-1 \right)\times 4 \\
 & \Rightarrow f'=f+55\times 4 \\
 & \Rightarrow f'=f+220 \\
\end{align}$
As it is given that the frequency of the last fork is $3$ times that of the first, so we will now put the value of the frequency of the last tuning fork as thrice the frequency of the first tuning fork. The expression will now look like this:
$\begin{align}
  & 3f=f+220 \\
 & \Rightarrow 3f-f=220 \\
 & \Rightarrow 2f=220 \\
 & \therefore f=110 \\
\end{align}$
The frequency of the first tuning fork came out to be $110$.

Hence the correct option will be $A$.

Additional information:
If some numbers are given in arithmetic progression, let us assume that the first term of the given arithmetic progression is $a$, the number of terms given is $n$ and the common difference between the terms of the given arithmetic progression is $d$, then the $n\text{th}$term of the arithmetic progression is given by:
${{a}_{n}}=a+\left( n-1 \right)d$
This same formula has been used to find out the frequency of the last tuning fork.

Note:
As the question says that each of the $56$ tuning forks give $4$ beats per second with the previous one, it means that each tuning fork gives a frequency of $4$ beats per second more as compared to the previous tuning fork and all of them hence form an arithmetic progression.