Arithmetic Progression Explained
An arithmetic progression is a sequence where the difference between consecutive terms is constant. This pattern appears frequently in real life and is ess
TL;DR: An arithmetic progression is a sequence where the difference between consecutive terms is constant. This pattern appears frequently in real life and i…
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Arithmetic Progression: An arithmetic progression is a sequence where the difference between consecutive terms is constant. This pattern appears frequently in real life and is ess
Arithmetic Progression Explained Simply
An arithmetic progression is a sequence where the difference between consecutive terms is constant. This pattern appears frequently in real life and is ess This is a key Mathematics concept for Class 10.
Definition and Common Difference
An arithmetic progression (AP) is a sequence where each term after the first is obtained by adding a fixed number to the previous term. The fixed number is called the common difference, denoted d. For example, 2, 5, 8, 11, 14 is an AP with first term a equals 2 and common difference d equals 3. The general form is a, a plus d, a plus 2d, a plus 3d, and so on. An AP can be increasing if d is positive, decreasing if d is negative, or constant if d equals 0.
General Term Formula
The nth term of an AP is given by the formula a_n equals a plus (n minus 1)d, where a is the first term, d is the common difference, and n is the term number. This formula allows you to find any term without listing all previous terms. For example, in the AP 2, 5, 8, 11, 14 where a equals 2 and d equals 3, the 10th term is a_10 equals 2 plus (10 minus 1)3 equals 2 plus 27 equals 29. You can also use this formula to find n if you know a_n.
Sum of Arithmetic Progression
The sum of the first n terms of an AP is given by S_n equals n divided by 2 times (2a plus (n minus 1)d), or equivalently S_n equals n divided by 2 times (first term plus last term). For example, to find the sum of the first 10 terms of 2, 5, 8, 11: S_10 equals 10 divided by 2 times (2(2) plus (10 minus 1)3) equals 5 times (4 plus 27) equals 5 times 31 equals 155. If the last term a_n is known, you can use S_n equals n divided by 2 times (a plus a_n).
Finding Terms and Common Difference
Given certain terms of an AP, you can find the first term and common difference. If you know the first and last terms and the number of terms, use the formulas above. If you know any two terms a_p and a_q, then a_q equals a_p plus (q minus p)d, so d equals (a_q minus a_p) divided by (q minus p). For example, if the 3rd term is 8 and the 7th term is 20, then 20 equals 8 plus (7 minus 3)d, so 4d equals 12 and d equals 3. Then the first term a equals a_3 minus 2d equals 8 minus 6 equals 2.
Applications of Arithmetic Progressions
APs model many real-world situations. If you save a fixed amount each month, your total savings form an AP. If a car depreciates by the same amount each year, its value forms a decreasing AP. Simple interest forms an AP where interest is added linearly. Building or construction projects with uniform increments often follow APs. Population growth at a constant rate and scheduled payment plans also form APs.
Worked Examples
Example 1. In an AP, the 5th term is 20 and the 10th term is 45. Find the first term and common difference.
Solution: Using a_n equals a plus (n minus 1)d: a_5 equals a plus 4d equals 20 and a_10 equals a plus 9d equals 45. Subtracting: (a plus 9d) minus (a plus 4d) equals 45 minus 20, so 5d equals 25 and d equals 5. Then a equals 20 minus 4(5) equals 0. The first term is 0 and common difference is 5.
Example 2. Find the sum of the first 20 terms of the AP 3, 7, 11, 15...
Solution: Here a equals 3, d equals 4. S_20 equals 20 divided by 2 times (2(3) plus (20 minus 1)4) equals 10 times (6 plus 76) equals 10 times 82 equals 820.
Example 3. How many terms of the AP 2, 5, 8, 11... give a sum of 155?
Solution: a equals 2, d equals 3. S_n equals n divided by 2 times (2a plus (n minus 1)d) equals 155. n divided by 2 times (4 plus 3n minus 3) equals 155. n divided by 2 times (1 plus 3n) equals 155. n plus 3n^2 equals 310. 3n^2 plus n minus 310 equals 0. Using the quadratic formula or factoring: (3n plus 31)(n minus 10) equals 0. n equals 10 (n must be positive). The first 10 terms sum to 155.
Where You See This in Real Life
- An employee gets a fixed salary increase of 1000 rupees each year, forming an AP of annual salaries.
- A child's height increases by approximately 5 cm per year, forming an AP of heights until growth stops.
- Depreciation of a vehicle by a fixed amount each year creates an AP of car values over time.
- Street numbering on one side often forms an AP: odd numbers 1, 3, 5, 7 with common difference 2.
Common Mistakes Students Make
- Confusing the term number with the term value - the nth term is a_n, not n
- Using the wrong formula for common difference - it is the difference between consecutive terms
- Forgetting that the first term is a_1, not a_0 - n starts at 1
- Making arithmetic errors in sum formula - be careful with parentheses and order of operations
Frequently Asked Questions
What is the difference between an AP and a geometric progression?
In an AP, consecutive terms have a constant difference. In a geometric progression, consecutive terms have a constant ratio. AP: 2, 5, 8, 11 (add 3 each time). GP: 2, 6, 18, 54 (multiply by 3 each time).
Can the common difference be negative or zero?
Yes, the common difference d can be any real number. If d is negative, the AP is decreasing. If d equals 0, all terms are the same constant value. If d is positive, the AP is increasing.
How is the sum formula derived?
Write the sum forward and backward, then add them: S equals a plus (a + d) plus ... plus l. S equals l plus (l minus d) plus ... plus a. Adding gives 2S equals n(a plus l), so S equals n(a plus l) divided by 2, where l is the last term.