Question

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. The Indian boxer delivers 8 punches in the first round and his punch delivery increases by 2 in each successive round, now if 1 punch delivered to 1 point, then who will win a gold medal.

Answer 1

In this particular question use the concept of arithmetic progression and the formula of the sum of an A.P series is given as ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where symbols have their usual meanings so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
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 ${13^{th}}$ round.
Now as we see that the above series forms an A.P, with first term, a = 12, common difference, $d = 13 – 12 = 14 – 13 = 1$, and the number of terms, $n = 13$.
So the sum of the series is
$\Rightarrow {S_n} = 12 + 13 + 14 + ........{\text{till 1}}{{\text{3}}^{th}}{\text{ term}}$
Now as we know that the sum of an A.P series is given as ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where symbols have their usual meanings.
Now substitute the values we have,
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {2\left( {12} \right) + \left( {13 - 1} \right)1} \right)$
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {24 + 12} \right) = 13\left( {12 + 6} \right) = 13\left( {18} \right) = 234$ Punches.
Now one punch = 1 point.
So 234 punches = 234 points.
Now the Indian boxer delivers 8 punches in the first round and his punch delivery increases by 2 in each successive round.
Let the first term of Indian boxer is 8, and after that his punches delivery increases by 2, so the next term is 10, thereafter 12, and so on.............
So the series is $8, 10, 12.............$ till ${13^{th}}$ round.
Now as we see that the above series forms an A.P, with first term, a = 8, common difference, $d = 10 – 8 = 12 – 10 = 2$, and the number of terms, $n = 13$.
So the sum of the series is
$\Rightarrow {S_n} = 8 + 10 + 12 + ........{\text{till 1}}{{\text{3}}^{th}}{\text{ term}}$
Now as we know that the sum of an A.P series is given as ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where symbols have their usual meanings.
Now substitute the values we have,
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {2\left( 8 \right) + \left( {13 - 1} \right)2} \right)$
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left( {16 + 24} \right) = 13\left( {8 + 12} \right) = 13\left( {20} \right) = 260$ Punches.
Now one punch = 1 point.
So 260 punches = 260 points.
Now as we see that Indian boxers earn 260 points whereas Chinese boxers earn 234 points.
So Indian boxer wins a Gold medal.
So this is the required answer.

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 whoever gets the highest points wins the gold medal.