Sequences and Series

2.1 Core concepts

• A sequence is an ordered list of numbers a_1, a_2, a_3, …, a_n, …; subscripts denote position and a_n is called the general term or nth term (NCERT §8.2, p. 137). The word "sequence" emphasises order; rearranging the terms generally produces a different sequence.
• A sequence with a finite number of terms is a finite sequence; otherwise it is an infinite sequence. Example: ancestors over 10 generations 2, 4, 8, …, 1024 is finite; the successive quotients 3, 3.3, 3.33, … in dividing 10 by 3 form an infinite sequence (NCERT §8.2, p. 136–137).
• Terms of a sequence need not follow a closed formula. Sometimes they are defined by a recurrence relation, e.g. the Fibonacci sequence a_1 = a_2 = 1, a_n = a_(n−1) + a_(n−2) for n > 2; primes 2, 3, 5, 7, … have no formula at all (NCERT §8.2, p. 136–137). Both are valid sequences's sense.
• A sequence can be regarded as a function whose domain is the set of natural numbers (or a subset of it); sometimes a(n) is used in place of a_n (NCERT §8.2, p. 137). This functional view is what makes sequences amenable to calculus (limits, convergence) in Class XII and beyond.
• A series associated with a sequence a_1, a_2, …, a_n, … is the expression a_1 + a_2 + … + a_n + …. It is finite or infinite according to the parent sequence, and is written compactly as Σ_(k=1)^n a_k using the Greek letter sigma (NCERT §8.3, p. 137).
• The word series refers to the indicated sum, not to the sum itself; for example 1 + 3 + 5 + 7 is a four-term series whose sum is 16 (NCERT §8.3, Remark, p. 137). CUET sometimes asks "write the series" — meaning write the indicated sum, not just the value.
• A sequence a_1, a_2, … is a Geometric Progression (G.P.) if each term is non-zero and a_(k+1)/a_k = r (constant) for k ≥ 1; a is the first term and r the common ratio, so the GP is a, ar, ar², ar³, … (NCERT §8.4, p. 139). The common ratio can be positive, negative, or a non-zero fraction, but never 0 (else the sequence terminates).
• General term of a GP: a_n = ar^(n−1). A finite GP is a, ar, …, ar^(n−1); an infinite GP is a, ar, ar², …, ar^(n−1), … (NCERT §8.4.1, p. 140). The exponent of r is n − 1, not n — the single most common student error.
• Sum to n terms of a GP: If r = 1, S_n = na. If r ≠ 1, S_n = a(r^n − 1)/(r − 1), equivalently a(1 − r^n)/(1 − r), derived by subtracting rS_n from S_n (NCERT §8.4.2, p. 140). The "easier" form depends on whether |r| > 1 or |r| < 1; both are algebraically identical.
• Standard worked examples that CUET mirrors: ancestors over 10 generations 2 + 4 + 8 + … + 2¹⁰ = 2046 (Example 11, p. 143); bacteria doubling every hour; quarterly bank interest. The pattern is always to identify a, r, n and plug into the sum formula.
• Geometric Mean (G.M.) of two positive numbers a and b is √(ab). Three numbers a, G, b are in GP iff G = √(ab); e.g. G.M. of 2 and 8 is 4, and 2, 4, 8 is a GP (NCERT §8.4.3, p. 143). G.M. is undefined (or complex) if either of a, b is negative.
• Inserting n GMs between a and b: If G_1, G_2, …, G_n lie between positive numbers a and b so that a, G_1, G_2, …, G_n, b is a GP, then b is the (n+2)th term: b = ar^(n+1), giving r = (b/a)^(1/(n+1)); the GMs are G_k = a(b/a)^(k/(n+1)) (NCERT §8.4.3, p. 143). The total number of terms is n + 2 because a and b are also part of the GP.
• Relationship A.M. ≥ G.M.: For two positive numbers a, b, let A = (a+b)/2 and G = √(ab). Then A − G = (a + b − 2√(ab))/2 = (√a − √b)²/2 ≥ 0, hence A ≥ G (NCERT §8.5, p. 144). Equality holds iff √a = √b, i.e. a = b. The inequality generalises to n positive numbers (n-variable AM-GM), but the NCERT chapter restricts to two.
• Arithmetic-progression material (covered in Class X) and the convergence of infinite GPs are out of scope here (sum-to-infinity S∞ = a/(1 − r) for |r| < 1 is in the older syllabus and may appear as Misc. Exercise but is not central in the 2026-27 edition).

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Portrait of Fibonacci (1175–1250) appears alongside the introduction (NCERT §8.1, p. 135). Sequences and series model real-world growth and decay despite their abstract appearance.
• Two-step derivation of the GP sum (NCERT §8.4.2, p. 140): (i) write S_n = a + ar + … + ar^(n−1); (ii) multiply by r to get rS_n = ar + ar² + … + ar^n; subtract the first from the second to obtain (r − 1)S_n = a(r^n − 1). Memorise the derivation: CUET sometimes asks for a sketch of it as a statement-correctness MCQ.
• The algebraic identity A − G = (√a − √b)²/2, which makes the AM–GM inequality visible at one glance (NCERT §8.5, p. 144). The square is non-negative ⇒ A − G ≥ 0 ⇒ A ≥ G.
• Worked Example 11 (ancestors): a = 2, r = 2, n = 10, so S_10 = 2(2¹⁰ − 1) = 2046, illustrating a real-life GP sum (NCERT §8.4.2, p. 143). This is the prototype for every "doubling over n periods" question.
• Process — identify GP, then plug: (i) Read first term a and common ratio r from the sequence. (ii) Decide whether the question asks for a_n (use general term) or S_n (use sum formula). (iii) Substitute and simplify; check whether r = 1 to avoid division by zero.
• Process — insert n GMs: (i) Treat a and b as the first and (n+2)-th terms of a GP with n+1 ratios in between. (ii) Compute r = (b/a)^(1/(n+1)). (iii) The k-th GM is a · r^k.
• Process — find two numbers from A.M., G.M.: Given A and G, use a + b = 2A and ab = G². Then a, b are roots of t² − 2A t + G² = 0. Discriminant 4A² − 4G² ≥ 0 is guaranteed by AM-GM.

2.4 Common confusions / NTA trap points

• Sigma notation vs. value. A series is the indicated sum, not the number that results. 1 + 3 + 5 + 7 is the series; 16 is its sum (NCERT §8.3 Remark, p. 137). NTA exploits this by asking what the "series" is.
• nth term off-by-one. In a GP, a_n = ar^(n−1), not ar^n. Students who write a_10 = ar^10 lose marks on direct questions (NCERT §8.4.1, p. 140).
• Wrong sum formula when r = 1. S_n = a(r^n − 1)/(r − 1) is undefined when r = 1; in that case S_n = na (NCERT §8.4.2, p. 140). A common trap option is n·a paired with the r ≠ 1 formula in the same MCQ.
• G.M. sign. G.M. of two positive numbers is the positive square root √(ab) (NCERT §8.4.3, p. 143). When inserting GMs the common ratio may admit two real values (e.g. r = ±4 between 1 and 256), but the definition of GM picks the positive root.
• A.M. ≥ G.M. holds only for positive numbers. The proof relies on (√a − √b)², which needs a, b > 0 (NCERT §8.5, p. 144). For mixed-sign numbers the inequality may reverse or be undefined.
• GP requires non-zero terms. A sequence with a zero term is excluded from being a GP because the ratio a_(k+1)/a_k would be undefined (NCERT §8.4, p. 139).
• Confusing AP and GP general terms. AP: a_n = a + (n − 1)d. GP: a_n = a · r^(n−1). The structural difference is addition vs. multiplication.
• Mis-counting inserted means. Inserting n GMs between a and b gives a GP with n + 2 terms total. Many students treat the count as n + 1.
• Mistaking G.M. for the average of GMs. G.M. of a and b is √(ab), not (√a + √b)/2.
• Mis-applying sum formula to infinite GP. The finite-sum formula S_n is unchanged; the limit as n → ∞ requires |r| < 1 and is a/(1 − r).
• Forgetting domain restrictions for n. n is a positive integer; sums like S_0 or S_{−1} are not defined.
• Misreading common ratio. In 8, 4, 2, 1, …, r = 1/2, not 2. Read direction carefully.
• Confusing exponent base. a · 4^(n−1) = … and a · 2^(n−1) = … produce different n's; identify r correctly before solving.
• Treating G.M. as signed. Even if a, b satisfy ab > 0 while both being negative, G.M. is defined in NCERT only for positive a, b.
• Off-by-one in inserting means. Common-ratio formula r = (b/a)^(1/(n+1)) has denominator n + 1, not n — corresponding to the n + 1 gaps between n + 2 terms.
• Equating sum of GP to product. S_n is a sum, not a product of terms; the product of n GP terms is (a · ar^(n−1))^(n/2) (geometric structure, not in the syllabus but a common distractor).

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 4, p. 140). Find the 10th term of the GP 5, 25, 125, …

Step 1 — identify a, r: a = 5, r = 25/5 = 5. Step 2 — apply general term: a_10 = a · r⁹ = 5 · 5⁹ = 5¹⁰. Step 3 — answer: 5¹⁰ (≈ 9 765 625).

Example B (NCERT Example 5, p. 141). Which term of the GP 2, 8, 32, … equals 131072?

Step 1 — identify a, r: a = 2, r = 8/2 = 4. Step 2 — set up equation: 2 · 4^(n−1) = 131072 ⇒ 4^(n−1) = 65536. Step 3 — solve: 4⁸ = 65536, so n − 1 = 8 ⇒ n = 9. Answer: 9th term.

Example C (NCERT Example 11, p. 143). Find the sum of ancestors over 10 generations (parents, grandparents, …).

Step 1 — model: GP with a = 2 (parents), r = 2 (each generation doubles), n = 10. Step 2 — apply sum formula: S_10 = 2(2¹⁰ − 1)/(2 − 1) = 2(1024 − 1) = 2 · 1023. Step 3 — answer: S_10 = 2046 ancestors.

Example D (NCERT Example 12, p. 144). Insert three numbers between 1 and 256 to form a GP with positive ratio.

Step 1 — model: 1, G_1, G_2, G_3, 256 is a 5-term GP, so 256 = 1 · r⁴. Step 2 — solve for r: r⁴ = 256 ⇒ r = 4 (positive root). Step 3 — compute means: G_1 = 4, G_2 = 16, G_3 = 64. Answer: 4, 16, 64.

Example E (NCERT Example 13, p. 144–145). Find two positive numbers whose A.M. is 10 and G.M. is 8.

Step 1 — write system: a + b = 20; ab = 64. Step 2 — form quadratic: a, b are roots of t² − 20t + 64 = 0; discriminant = 400 − 256 = 144. Step 3 — solve: t = (20 ± 12)/2 = 16 or 4. Answer: 4 and 16.