India’s Most Loved Science & Maths Videos – Packed in a Mobile App!
Introducing Avanti Gurukul App
Get Download Link on your Phone
India’s Most Loved Science & Maths Videos – Packed in a Mobile App!
Introducing Avanti Gurukul App
Arithmetic Progression Class 10 – Continuing our series for Class 10 Revision for CBSE board exams, we bring you a new chapter today i.e. Arithmetic Progression
All the questions in this chapter are based on basic definition of AP and 2 equations that are derived from it. Concepts to be covered in this article are:
- What is an Arithmetic Progression ?
- nth term of an Arithmetic Progression
- Sum of first n terms of an Arithmetic Progression
Let’s discuss each topic of Arithmetic Progression one-by-one
What is an AP?
An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number ? to the preceding term, except the first term. The fixed number ? is called the common difference. This means, that we start with any number, say ‘?’ and keep adding ‘?’ to it to obtain the subsequent terms.
So, any AP can be written as: ?, ?+?, ?+2? and so on.
Remember that the difference between 2 consecutive terms of AP will always remain same, i.e.,
nth term of an Arithmetic Progression
The nth term of an AP is given by a(n)= a + (n-1)d
Where a is the first term of AP and d is the common difference.
This is the most basic equation of AP and should be remembered at all times. Using this equation, we can find any one variable, i.e., nth term, first term, number of terms or common difference, if remaining terms are given.
Sum of first n terms of an Arithmetic Progression
The sum of first ? terms of an AP is given by:
S(n)= (n/2)[2a + (n-1)d]
or S(n)= (n/2)[a+l]
Where l= a + (n-1)d = nth term or last term of AP
Practice problems – Arithmetic Progression
Let’s solve a few questions that have been asked in previous boards paper:
Q5. Reshma wanted to save at least ₹ 6,500 for sending her daughter to school next year (after 12 months). She saved ₹ 450 in the first month and raised her saving by ₹ 20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year? What value is reflected in this question? 4 Marks, CBSE 2016, Foreign (30/2/1)
Sol: 1st month : ₹ 450
2nd month: ₹ (450+20)=₹ 470 ….
This will form an AP with: ?=450, ?=20 and ?=12
?(12)=(12/2)[2×450+(12−1)20]=6720
⇒ Savings at the end of 1 year =₹ 6720 (>₹ 6500)
∴ She will be able to send her daughter to school.
Value: Small savings can fulfill big dreams.