Question

If a clock strikes once at 1 o'clock, twice at 2 o'clock and so on, how many times will it strike in a day?

Answer 1

Hint: The total number of times the clock strikes in a day will be the sum of the number of times it strikes at each hour. Thus, while evaluating the total, the sum is in the form of an arithmetic progression and thus can be found out from the sum of arithmetic progression formulas.

Complete step-by-step solution:

It is given that the clock strikes once at 1 o'clock, twice at 2 o'clock etc. Thus, we can state this as

The number of times the clock strikes at n o'clock=n ……………..(1.1)

Let the total number of times the clock strikes in a day be given by S. Then, S will be the sum of the number of times the clock strikes at 1 o'clock, the number of times the clock strikes at 3 o'clock etc. upto 24 hours as there are 24 hours in a day (the clock would complete 12 hours from 1o'clock to 12 o'clock and again come to 1 o'clock to complete 24 hours)

Thus, S= (number of times it struck at 1 o'clock+ number of times it struck at 2 o'clock + …+ number of times it struck at 12 o'clock) + (number of times it struck at 1 o'clock+…+ number of times it struck at 12 o'clock)

(as the clock would complete two circles in 24hours).

$\Rightarrow S=\text{ }\left( 1+2+\ldots +12 \right)\text{ }+\text{ }\left( 1+2+\ldots +12 \right)\text{ }\ldots \ldots \ldots .\left( 1.2 \right)$

The sum of an arithmetic progression with first term a1 and common difference d upto n terms is given by 

${{s}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+(n-1)d \right)\text{ }.............(1.3)$

Thus taking $a_1$=1 and d=1 and n=12 in equation (1.3), we obtain

\[1+2+\ldots .+12=\dfrac{12}{2}\left( 2\times 1+(12-1)\times 1 \right)=\dfrac{12}{2}\left( 2+11 \right)=\dfrac{12\times 13}{2}\]

Thus, using this in equation (1.2), we get

$S=\dfrac{12\times 13}{2}+\dfrac{12\times 13}{2}=12\times 13=156$

Thus, the clock will strike 156 times in a day.

Note: In the calculation it is important to consider that in 24 hours, the clock would make two full rounds from 1 to 12 and not just one round from 1 to 24. Thus, the sum will be in the form of two arithmetic progressions from 1 to 12 with common difference 1.