Question
Question: In a 13 round boxing match, a boxer from India and from China competed for the gold medal. The Chine...
In a 13 round boxing match, a boxer from India and from China competed for the gold medal. The Chinese boxer delivers 12 punches in the first round and his punches delivery increases by 1 in each round. Find the total number of punches delivered by the chinese boxer.
Solution
In this particular question use the concept of arithmetic progression and the formula of the sum of an A.P series is given as Sn=2n[2a+(n−1)d] where symbols have their usual meanings so use these concepts to reach the solution of the question.
Formula used: The formula of sum of an A.P series is given as
Sn=2n[2a+(n−1)d]
Where a is the first term,
d is the common difference,
n is the number of terms.
Complete step-by-step solution:
Given: - There is a 13 round boxing match, where boxers from India and from China compete for the gold medal.
The Chinese boxer delivers 12 punches in the first round and his punches delivery increases by 1 in each round.
Let the first term of a Chinese boxer is 12, and after that his punches delivery increases by 1, so the next term is 13, thereafter 14, and so on.
So, the series is 12,13,14,… till 13th round.
Now as we see that the above series forms an A.P, with
⇒ First-term, a=12
⇒ Common difference, d=13−12=1
⇒ Number of terms, n=13
So, the sum of the series is,
⇒S13=213[2×12+(13−1)×1]
Subtract the value in the inner bracket,
⇒S13=213[2×12+12×1]
Now, multiply the terms,
⇒S13=213[24+12]
Now add the terms,
⇒S13=213×36
Cancel out the common factor,
⇒S13=13×18
Now, multiply the terms,
∴S13=234
Hence, the number of punches delivered by the Chinese boxer is 234 punches.
Note: Whenever we face such types of questions always solved by using the arithmetic progression, so first find out the series and the values of the first term, common difference and the number of terms as above then substitute these values in the formula of the sum as above and simplify the expression.