Question

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 $I$ will plant $1$ tree, a section of class $II$ will plant $2$ trees and so on till Class $XII$. There are three sections in each class. How many trees will be planted by the students

Answer 1

Given:

The number of trees that each section of each class will plant will be same as the class in which they are studying.

Number of classes are $12$ ($I$ to $XII$), each class has $3$ sections.

$\Rightarrow$ Number of trees planted by the students of each section of class $I=1\times 3=3$ [Since, each class has three sections] ---(1)

Number of trees planted by the students of each section of class $II=2\times 3=6$---(2)

Number of trees planted by the students of each section of class $=3\times 3=9$---(3)

So, the number of trees planted by the students of different classes are,

$3+6+9+.....$ up to $12$ terms.

Here, $6-3=9-6=...=3$

Since, the difference of every consecutive terms is constant.

So, this series is an arithmetic series.

It is clear that $a=3,d=6-3=3$ and $n=12$

Since, the sum of first $n$-terms of AP is,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$\Rightarrow S_{12}=\frac{12}{2}[6+(3\times 11\left)\right]$

$\Rightarrow S_{12}=6[6+33]$

$\Rightarrow S_{12}=6\left[39\right]$

$\Rightarrow S_{12}=234$

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

(OR)

Take, $3+6+9+.....36$ [Since, the last class is $12$ and $12\times 3=36$]

Now, $a=3,d=6-3=3,n=12$ and, Last term $a_n=36$

Since, the sum of first $n$ -terms of AP is,

$S_n=\frac{n}{2}[a+a_n]$

$\Rightarrow S_{12}=\frac{12}{2}[3+36]$

$\Rightarrow S_{12}=6\times 39$

$\Rightarrow S_{12}=234$

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