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Chapter 5 : Arithmetic Progression | 10th class board | cbse board


Chapter 5 : Arithmetic Progression

Master Arithmetic Progressions (AP) | Class 10 Math Guide

Have you ever noticed the spiral pattern on a pineapple, the petals of a sunflower, or the decreasing lengths of the rungs on a ladder? Mathematics is the hidden language behind these natural and man-made patterns.
In Class 10, one of the most fascinating patterns we study is the Arithmetic Progression (AP). Whether you are figuring out a monthly savings plan or calculating compound interest, AP is an incredibly handy tool. Let’s break down the theory and tackle some practice questions!


What is an Arithmetic Progression?

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
Common Difference (d): This fixed number that we add is called the common difference of the AP. Remember, d can be positive, negative, or zero.
First Term (a): The starting number of your sequence.

The General Form of an AP

If a is your first term and d is your common difference, the general form of an AP looks like this:
a, a+d, a+2d, a+3d, ...
Example:
If your first term a = 6 and common difference d = 3, your AP is:
6, 9, 12, 15...


Finding the nth Term of an AP

Imagine you are offered a job with a starting salary of ₹8000 and an annual increment of ₹500. How would you calculate your salary for the 15th year without adding 500 fourteen times?
You use the formula for the nth term (or general term) of an AP:

an = a + (n-1)d

where an = the nth term you want to find
a = the first term
n = the position of the term
d = the common difference
(Note: If there are m terms in the AP, then am represents the last term, which is sometimes denoted by l.)


Sum of the First n Terms of an AP

What if you save ₹100 on your first birthday, ₹150 on your second, ₹200 on your third, and so on? How much money will you have collected by your 21st birthday? Adding them one by one is too tedious.Instead, use the sum formula:

Alternatively, if you already know the first term (a) and the last term (l), you can use this shortcut:


Practice Question with solution

Question 1: Identifying an AP
List A: 4, 10, 16, 22, ...
Solution:
Step 1: Find the difference between consecutive terms.
10 - 4 = 6
16 - 10 = 6
22 - 16 = 6
Conclusion: Since the difference is a constant 6, this is an AP with a common difference d = 6.
Next two terms: 22 + 6 = 28,
and 28 + 6 = 34.
List B: 1, 1, 1, 2, 2, 2, 3, 3, ...
Solution:
Step 1: Check the differences.
1 - 1 = 0
2 - 1 = 1
Conclusion: The difference between terms is not constant. Therefore, this is not an AP.


Question 2: Finding a Specific Term : Find the 10th term of the AP: 2, 7, 12, ...
Step 1: Identify given values.
First term (a) = 2
Common difference (d) = 7 - 2 = 5
Position (n) = 10
Step 2: Apply the nth term formula.
an = a + (n-1)d
a10 = 2 + (10 - 1)5
a10 = 2 + (9 * 5)
a10 = 2 + 45 = 47
Answer: The 10th term is 47.


Question 3: Finding the Number of Terms : Which term of the AP: 21, 18, 15, ... is -81?
Solution:
Step 1: Identify given values.
First term (a) = 21
Common difference (d) = 18 - 21 = -3
nth term (an) = -81
Step 2: Apply the formula and solve for n.
an = a + (n-1)d
-81 = 21 + (n - 1)(-3)
-81 - 21 = -3(n - 1)
-102 = -3(n - 1)
Divide both sides by -3
34 = n - 1
n = 35
Answer: -81 is the 35th term of the AP.


Question 4: Sum of an AP : How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?
Solution
Step 1: Identify given values.
a = 24
d = 21 - 24 = -3
Sum (Sn) = 78
Step 2: Apply the sum formula.
Sn = n/2[2a + (n-1)d]
78 = n/2[2(24) + (n-1)(-3)]
156 = n[48 - 3n + 3]
156 = n[51 - 3n]
156 = 51n - 3n2.
Step 3: Solve the quadratic equation.
Rearrange to standard form: 3n2 - 51n + 156 = 0
Divide the entire equation by 3 to simplify:
n2 - 17n + 52 = 0
Factor the equation: (n - 4)(n - 13) = 0
n = 4 or n = 13
Answer: The number of terms can be either 4 or 13.

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