AP — nth term, sum, AM
Sequences & Series
14
JEE Qs
8%
Hard
75
min
Master the fundamental definitions, formulas, and properties of AP, especially the relationship between a_n and S_n, and strategic term selection to efficiently solve problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1The nth term of an AP (a_n) is a linear function of n (i.e., of the form An + B), where A is the common difference and B = a-d.
- 2The sum of n terms of an AP (S_n) is a quadratic function of n without a constant term (i.e., of the form An^2 + Bn), where A = d/2 and B = a - d/2.
- 3The nth term can be derived from the sum of n terms using the relation a_n = S_n - S_{n-1} (for n > 1) and a_1 = S_1. This is crucial for problems where S_n is given as a function of n.
- 4For problems involving a specific number of terms whose sum is given (e.g., 3, 4, or 5 terms), choose the terms strategically to simplify calculations: for 3 terms (a-d, a, a+d); for 4 terms (a-3d, a-d, a+d, a+3d); for 5 terms (a-2d, a-d, a, a+d, a+2d).
- 5If three numbers a, b, c are in AP, then the middle term b is the arithmetic mean of a and c, implying 2b = a+c.
⚠️ Common Mistakes
- ✕Off-by-one errors when calculating the number of terms, especially when the sequence does not start from the first term or when inserting 'm' arithmetic means (m+2 terms in total).
- ✕Confusing the formula for the nth term (a_n) with the sum of n terms (S_n), or misinterpreting what each formula calculates.
- ✕Incorrectly applying the relationship a_n = S_n - S_{n-1} for n=1, as S_0 is undefined (always ensure n > 1 for S_n - S_{n-1}).
📝 Practice Questions
See allQ12.The remainder, when 7103 is divided by 23 , is equal to : (1) 6 (2) 17 (3) 9 (4) 14
Q22.Let a1, a2, … , a2024 be an Arithmetic Progression such that a1 + (a5 + a10 + a15 + … + a2020) + a2024 = 2233. Then a1 + a2 + a3 + … + a2024 is equal to _______ 1 2 3 , then α is equal to ________ (3x + t = 5eα ( 85 )
Q1. Let a1, a2, a3, … be a G.P. of increasing positive terms. If a1a5 = 28 and a2 + a4 = 29, then a6 is equal to: (1) 628 (2) 812 (3) 526 (4) 784 = 0. If x(1) = 1, then x ( 12 ) is :
Q15.If ∑nr=1 Tr = (2n−1)(2n+1)(2n+3)(2n+5)64 , then limn→∞∑nr=1 ( Tr1 ) (1) 0 (2) 23 (3) 1 (4) 13
Q13.Suppose that the number of terms in an A.P. is 2k, k ∈N . If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : (1) 6 (2) 5 (3) 8 (4) 4 y+2
Q1. If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to (1) −1080 (2) −1020 (3) −1200 (4) −120
NCERT Chapters
- Class 11 Mathematics Ch 9: Sequences and Series