An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This constant is called the common difference (d). APs appear naturally in problems about savings, seating arrangements, salary increments, and physical patterns.
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Key Concepts
- Definition: A sequence a, a+d, a+2d, a+3d, ... is an AP with first term a and common difference d.
- d can be positive, negative, or zero.
- Example: 2, 5, 8, 11, ... is an AP with a = 2, d = 3.
General (nth) Term:
an = a + (n - 1)d
Sum of First n Terms:
Sn = n/2 x [2a + (n-1)d] OR Sn = n/2 x (a + l), where l = last term.
Important Relationship: an = Sn - Sn-1 for n ≥ 2.
Checking if a sequence is an AP: Compute the difference between consecutive terms. If constant, it's an AP.
Three terms in AP: If a-d, a, a+d are in AP, their sum = 3a and product = a(a2 - d2). Using this form simplifies many problems.
Four terms in AP: Use a-3d, a-d, a+d, a+3d.
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Worked Examples
Find the 20th term of the AP: 3, 7, 11, 15, ...
a = 3, d = 4, n = 20.
a20 = 3 + (20-1) x 4 = 3 + 76 = 79.
How many terms are in the AP 7, 13, 19, ..., 205?
a = 7, d = 6, an = 205.
205 = 7 + (n-1) x 6. 198 = (n-1) x 6. n-1 = 33. n = 34.
Find the sum of first 25 terms of the AP 5, 8, 11, ...
a = 5, d = 3, n = 25.
S25 = 25/2 x [2(5) + 24(3)] = 25/2 x [10 + 72] = 25/2 x 82 = 25 x 41 = 1025.
The sum of first n terms of an AP is 5n2 + 3n. Find the nth term.
an = Sn - Sn-1 = (5n2 + 3n) - (5(n-1)2 + 3(n-1))
= 5n2 + 3n - 5n2 + 10n - 5 - 3n + 3
= 10n - 2.
an = 10n - 2. (Check: a1 = S1 = 5+3 = 8 = 10(1)-2 = 8. Correct!)
The 8th term of an AP is 37 and the 12th term is 57. Find the AP.
a + 7d = 37 ...(1); a + 11d = 57 ...(2).
Subtract (1) from (2): 4d = 20, d = 5.
From (1): a = 37 - 35 = 2. AP: 2, 7, 12, 17, ...
Three numbers are in AP. Their sum is 21 and product is 231. Find them.
Let numbers be a-d, a, a+d. Sum = 3a = 21, so a = 7.
Product = 7(7-d)(7+d) = 7(49 - d2) = 231. 49 - d2 = 33. d2 = 16. d = ±4.
Numbers: 3, 7, 11 (or 11, 7, 3).
How many multiples of 4 lie between 10 and 250?
First multiple of 4 ≥ 10 is 12. Last multiple ≤ 250 is 248.
a = 12, d = 4, an = 248. n = (248-12)/4 + 1 = 236/4 + 1 = 59 + 1 = 60.
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Key Formulas
Key formulas
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Common mistakes
- Forgetting the formula is a + (n-1)d, not a + nd.
- In Sn = n/2 [2a + (n-1)d], computing (n+1) instead of (n-1).
- When three terms are in AP, always use a-d, a, a+d (not a, a+d, a+2d) to simplify algebra.
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Summary
APs describe uniform change. The two core formulas — nth term and sum of n terms — together with the structure of terms in AP unlock a wide variety of examination problems. Practice identifying a and d quickly from given information.