Topics
Linear Equations in Two Variables
- Foundation of Linear Equations in Two Variables
- Simultaneous Linear Equations
- Graph of a Linear Equation in Two Variables
- Methods of Solving Linear Equations in Two Variables > Graphical Method
- Methods of Solving Linear Equations in Two Variables > Determinant method (Cramer’s Rule)
- Equations Reducible to Linear Equations
- Application of Simultaneous Equations
Quadratic Equations
- Concept of Quadratic Equations
- Factorisation Method
- Completing the Square Method
- Quadratic Formula (Shreedharacharya's Rule)
- Nature of Roots of a Quadratic Equation
- Relation Between Zeroes (Roots) and Coefficients of a Quadratic Equation
- Formation of a Quadratic Equation with Given Roots
- Application of Quadratic Equation
Arithmetic Progression
Financial Planning
Probability
Statistics
Introduction
Arithmetic Progressions are useful for describing many real-life situations where something increases or decreases uniformly, such as:
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Monthly savings
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Annual salary increments
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Yearly production of goods
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Loan instalments
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Distance covered in equal time intervals
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Arrangements of objects in rows
In this concept, you will learn how to identify an A.P. in word problems, find any required term, and calculate the total of several terms using the formulas for the nth term and the sum of n terms.
Common Clue Words
| Clue in question | Likely A.P. meaning |
|
"Increases by ₹500 every year"
|
Common difference d = 500
|
|
"Decreases by ₹1,500 every month"
|
Common difference d = -1500
|
|
"First month's saving is ₹200"
|
First term a = 200
|
|
"After how many months/years?"
|
Find n
|
|
"Total amount saved/paid/produced"
|
Find \[S_n\]
|
|
"Value in the 10th year"
|
Find \[a_{10}\]
|
Example 1
Question: Ajay sharma repays the borrowed amount of ₹ 3,25,000 by paying ₹ 30500 in the first month and then decreases the payment by ₹ 1500 every month. How long will it take to clear his amount?
Solution:
Step 1: Identify the given A.P. parameters
First term \[ a = 30{,}500 \]
Common difference \[ d = -1{,}500 \] (since payments decrease every month)
Total amount \[ S_{n} = 3{,}25{,}000 \]
Step 2: Substitute values into the sum formula ( \[ S_{n} \] )
\[ S_{n} = \frac{n}{2}[2a + (n - 1)d] \]
\[ 3{,}25{,}000 = \frac{n}{2}[2(30{,}500) + (n - 1)(-1500)] \]
Step 3: Form and solve the quadratic equation
Simplifying gives: \[ 3n^{2} - 125n + 1300 = 0 \]
Factorising: \[ (n - 20)(3n - 65) = 0 \]
\[ n = 20 \text{ or } n = \frac{65}{3} = 21\frac{2}{3} \]
Step 4: Select the valid number of months
Since \[ n \] must be a natural number, \[ n \neq \frac{65}{3}, \] therefore \[ n = 20. \]
Final Answer:
\[ \therefore \] It will take 20 months to clear the borrowed amount.
Example 2
Question: In the year 2010 in the village there were 4000 people who were literate. Every year the number of literate people increases by 400. How many people will be literate in the year 2020 ?
Solution:
Step 1: Identify the given parameters
First term \[ a = 4000. \]
Common difference \[ d = 400. \]
Total number of years \[ n = 11 \] (from 2010 to 2020).
Step 2: Apply the \[ n^{\text{th}} \] term formula ( \[ t_{n} = a + (n - 1)d \] )
\[ t_{11} = 4000 + (11 - 1) \times 400. \]
Step 3: Simplify and calculate
\[ t_{11} = 4000 + 4000 = 8000. \]
Final Answer:
In the year 2020, 8000 people will be literate.
