# In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying e.g., a section of Class ${\text{I}}$ will plant 1 tree, a section of Class ${\text{II}}$ will plant 2 trees and so on till Class ${\text{XII}}$. There are three sections of each class. How many trees will be planted by the students?

Last updated date: 27th Mar 2023

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Answer

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Hint- Convert the problem statement in the form of a series of $n$ natural numbers and then find the sum of first $n$ natural numbers.

According to the problem statement, every class has three sections.

Number of trees planted from Class ${\text{I}}$\[ = 3 \times 1\] (1 from each section)

Number of trees planted from Class \[{\text{II}}\]\[ = 3 \times 2\] (2 from each section)

Number of trees planted from Class ${\text{III}}$\[ = 3 \times 3\] (3 from each section)

Number of trees planted from Class ${\text{IV}}$\[ = 3 \times 4\] (4 from each section)

And so on up to Class\[{\text{XII}}\]

Number of trees planted from Class \[{\text{XII}}\]\[ = 3 \times 12\] (12 from each section)

Therefore, for the total number of trees planted we will add all the trees planted from all the classes.

Total number of trees planted\[ = 3 \times 1 + 3 \times 2 + 3 \times 3 + 3 \times 4 + ...... + 3 \times 11 + 3 \times 12 = 3\left[ {1 + 2 + 3 + 4 + ..... + 11 + 12} \right]\]

As we know that the sum of first $n$ natural numbers is given by \[{\text{S}} = \dfrac{{n\left( {n + 1} \right)}}{2}\]

Therefore, the sum of the first 12 natural numbers can be computed if we put \[n = 12\] in the above formula.

i.e., \[1 + 2 + 3 + 4 + ..... + 11 + 12 = \dfrac{{12\left( {12 + 1} \right)}}{2} = \dfrac{{12 \times 13}}{2} = 78\]

Total number of trees planted\[ = 3\left[ {1 + 2 + 3 + 4 + ..... + 11 + 12} \right] = 3 \times 78 = 234\]

Therefore, the total number of trees which are planted by the students are 234 trees.

Note- In these types of problems, it is very crucial to interpret the problem statement very carefully and missing any of the minute detail will lead to a wrong answer. If we observe this particular problem carefully, the sum is reduced to an arithmetic progression for which we have used directly the formula of the sum of first $n$ natural numbers because here the common difference is 1 for the series 1,2,3...12.

According to the problem statement, every class has three sections.

Number of trees planted from Class ${\text{I}}$\[ = 3 \times 1\] (1 from each section)

Number of trees planted from Class \[{\text{II}}\]\[ = 3 \times 2\] (2 from each section)

Number of trees planted from Class ${\text{III}}$\[ = 3 \times 3\] (3 from each section)

Number of trees planted from Class ${\text{IV}}$\[ = 3 \times 4\] (4 from each section)

And so on up to Class\[{\text{XII}}\]

Number of trees planted from Class \[{\text{XII}}\]\[ = 3 \times 12\] (12 from each section)

Therefore, for the total number of trees planted we will add all the trees planted from all the classes.

Total number of trees planted\[ = 3 \times 1 + 3 \times 2 + 3 \times 3 + 3 \times 4 + ...... + 3 \times 11 + 3 \times 12 = 3\left[ {1 + 2 + 3 + 4 + ..... + 11 + 12} \right]\]

As we know that the sum of first $n$ natural numbers is given by \[{\text{S}} = \dfrac{{n\left( {n + 1} \right)}}{2}\]

Therefore, the sum of the first 12 natural numbers can be computed if we put \[n = 12\] in the above formula.

i.e., \[1 + 2 + 3 + 4 + ..... + 11 + 12 = \dfrac{{12\left( {12 + 1} \right)}}{2} = \dfrac{{12 \times 13}}{2} = 78\]

Total number of trees planted\[ = 3\left[ {1 + 2 + 3 + 4 + ..... + 11 + 12} \right] = 3 \times 78 = 234\]

Therefore, the total number of trees which are planted by the students are 234 trees.

Note- In these types of problems, it is very crucial to interpret the problem statement very carefully and missing any of the minute detail will lead to a wrong answer. If we observe this particular problem carefully, the sum is reduced to an arithmetic progression for which we have used directly the formula of the sum of first $n$ natural numbers because here the common difference is 1 for the series 1,2,3...12.

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