Ex.5.3 Q17 Arithmetic Progressions Solution - NCERT Maths Class 10
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 of each class. How many trees will be planted by the students?
Text Solution
What is Known?
The number of trees planted by \(3\) sections of each class (I to XII).
What is Unknown?
Number of trees planted by the students.
Reasoning:
Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.
Steps:
Class I | Class II | Class III | …………………… | Class XII | |
Section \(A\) | \(1\) | \(2\) | \(3\) | …………………… | \(12\) |
Section \(B\) | \(1\) | \(2\) | \(3\) | …………………… | \(12\) |
Section \(C\) | \(1\) | \(2\) | \(3\) | …………………… | \(12\) |
Total | \(3\) | \(6\) | \(9\) | \(36\) |
It can be observed that the number of trees planted by the students are in an AP.
\[3, 6, 9, 12, 15,\dots\dots36\]
- First term \(a = 3\)
- Common difference, \(d = 6 - 3 = 3\)
- Number of terms, \(n = 12\)
We know that sum of \(n\) terms of AP,
\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{12}} &= \frac{{12}}{2}\left[ {2 \times 3 + \left( {12 - 1} \right) \times 3} \right]\\& = 6\left[ {6 + 11 \times 3} \right]\\ &= 6\left[ {6 + 33} \right]\\ &= 6 \times 39\\ &= 234\end{align}\]
Therefore, \(234\) trees will be planted by the students.