Sequences and series for NDA

The NDA syllabus lists "arithmetic, geometric and harmonic progressions" in a single line, and the questions that come from it are usually quick. You are given two terms and asked for a third, a sum and asked for a term, or two numbers and asked for their means. The formulas are few, but each has a condition that a hurried candidate forgets: r ≠ 1 for a GP sum, |r| < 1 for an infinite GP, positive numbers for AM ≥ GM.

Typical question types are:

• a term or a sum of an AP or GP from partial information;
• the number of terms in a given progression;
• the nth term when the sum of n terms is given as a formula;
• sums to infinity, including recurring decimals;
• AM, GM and HM of two numbers and the relations between them;
• standard sums such as 1² + 2² + … + n².

Arithmetic progression (AP)

An AP has a constant difference d between consecutive terms: a, a + d, a + 2d, …

• nth term: Tₙ = a + (n − 1)d
• sum of n terms: Sₙ = n/2 × [2a + (n − 1)d] = n/2 × (first term + last term)

Why the sum formula works. Write the sum forwards and backwards and add. Each pair (first + last, second + second-last, …) has the same total, and there are n pairs, so 2Sₙ = n(a + l).

Useful facts:

• a, b, c are in AP exactly when 2b = a + c; b is the arithmetic mean.
• For three unknown terms in AP, take a − d, a and a + d; the sum is simply 3a.
• If Sₙ is given, Tₙ = Sₙ − Sₙ₋₁ for n ≥ 2, and T₁ = S₁.

Geometric progression (GP)

A GP has a constant ratio r: a, ar, ar², …

• nth term: Tₙ = arⁿ⁻¹
• sum of n terms: Sₙ = a(rⁿ − 1)/(r − 1), valid for r ≠ 1 (if r = 1, Sₙ = na)
• sum to infinity: S = a/(1 − r), valid only when |r| < 1

Why the sum formula works. Multiply Sₙ by r and subtract: rSₙ − Sₙ = arⁿ − a, because every middle term cancels.

Useful facts:

• a, b, c are in GP exactly when b² = ac; b is the geometric mean.
• For three unknown terms in GP, take a/r, a and ar; the product is simply a³.
• A recurring decimal is an infinite GP: 0.454545… = 45/100 + 45/10,000 + … = (45/100) ÷ (1 − 1/100) = 45/99 = 5/11.

Harmonic progression (HP)

A sequence is an HP when the reciprocals of its terms are in AP. There is no direct sum formula. To find a term, flip to the AP, find the term there, and flip back.

• a, b, c are in HP exactly when b = 2ac/(a + c); b is the harmonic mean.

The three means

For two positive numbers a and b:

Two results link them:

• A ≥ G ≥ H, with equality only when a = b;
• G² = AH. For 4 and 9: 6.5 × 72/13 = 36 = 6².

AM ≥ GM also gives quick minimum values. For x > 0, x + 1/x ≥ 2√(x × 1/x) = 2, so the least value of x + 1/x is 2, at x = 1.

Standard sums

Worked NDA-style MCQs

Q1. The 10th term of an AP is 29 and the 20th term is 59. The 30th term is:
(a) 79 (b) 86 (c) 89 (d) 92

Subtracting, 10d = 30, so d = 3. From a + 9d = 29, a = 2. T₃₀ = 2 + 29 × 3 = 89. Answer: (c).

Q2. If the sum of the first n terms of a sequence is 3n² + 2n, its nth term is:
(a) 6n + 1 (b) 6n + 5 (c) 3n + 2 (d) 6n − 1

Tₙ = Sₙ − Sₙ₋₁ = (3n² + 2n) − [3(n − 1)² + 2(n − 1)] = 3n² + 2n − 3n² + 6n − 3 − 2n + 2 = 6n − 1. Check: T₁ = 5 = S₁. Answer: (d).

Q3. How many terms of 24, 21, 18, … must be taken for the sum to be 105?
(a) 7 only (b) 10 only (c) 7 or 10 (d) 8

n/2 × [48 + (n − 1)(−3)] = 105 gives n(51 − 3n) = 210, that is, n² − 17n + 70 = 0, so n = 7 or 10. Both are valid: the 8th, 9th and 10th terms are 3, 0 and −3, which add to 0. Answer: (c).

Q4. The fifth term of the HP 1/2, 1/5, 1/8, … is:
(a) 1/11 (b) 1/14 (c) 1/17 (d) 1/20

The reciprocals 2, 5, 8, … form an AP with d = 3. Its fifth term is 2 + 4 × 3 = 14, so the HP term is 1/14. Answer: (b).

Q5. The value of 0.272727… as a fraction is:
(a) 3/11 (b) 27/100 (c) 27/90 (d) 9/37

(27/100) ÷ (1 − 1/100) = 27/99 = 3/11. Answer: (a).

Q6. The AM of two positive numbers is 10 and their GM is 8. The numbers are:
(a) 12 and 8 (b) 18 and 2 (c) 14 and 6 (d) 16 and 4

a + b = 20 and ab = 64, so they are the roots of t² − 20t + 64 = 0, which are 16 and 4. Answer: (d).

Common mistakes

• Using a/(1 − r) when |r| ≥ 1. The series 1 + 2 + 4 + … has no finite sum.
• Counting terms wrongly. From 7 to 139 in steps of 4 there are (139 − 7)/4 + 1 terms. The "+ 1" is the most common slip in the chapter.
• Rejecting a valid answer. When a sum condition gives two values of n, check both before discarding one.
• Applying AM ≥ GM to negative numbers. It holds only for non-negative numbers.
• Adding HP terms like an AP. Only the reciprocals are in AP.

Practice set

1. The number of terms in the AP 7, 11, 15, …, 139 is: (a) 33 (b) 34 (c) 35 (d) 36
2. The sum of all two-digit multiples of 3 is: (a) 1,599 (b) 1,683 (c) 1,650 (d) 1,665
3. In a GP the 3rd term is 12 and the 6th term is 96. The 8th term is: (a) 192 (b) 256 (c) 384 (d) 768
4. The sum to infinity of 6 − 2 + 2/3 − … is: (a) 4.5 (b) 9 (c) 3 (d) 4
5. The sum of the first 10 terms of 2, 4, 8, … is: (a) 1,022 (b) 1,024 (c) 2,046 (d) 2,048
6. 1³ + 2³ + … + 10³ equals: (a) 385 (b) 55 (c) 2,025 (d) 3,025
7. The GM of 9 and 25 is: (a) 17 (b) 16 (c) 15 (d) 225
8. The least value of 4x + 9/x for x > 0 is: (a) 6 (b) 12 (c) 13 (d) 36

Answers:

1. (b) 34. (139 − 7)/4 + 1 = 33 + 1.
2. (d) 1,665. Terms from 12 to 99: (99 − 12)/3 + 1 = 30 terms; sum = 30/2 × (12 + 99) = 15 × 111.
3. (c) 384. r³ = 96/12 = 8, so r = 2 and a = 12/4 = 3; T₈ = 3 × 2⁷.
4. (a) 4.5. r = −1/3, so S = 6 ÷ (4/3).
5. (c) 2,046. 2(2¹⁰ − 1)/(2 − 1) = 2 × 1,023.
6. (d) 3,025. 55².
7. (c) 15. √225.
8. (b) 12. AM ≥ GM gives 4x + 9/x ≥ 2√36 = 12, with equality at x = 3/2.

What to do next

• Write every formula in this guide with its condition (r ≠ 1, |r| < 1, positive numbers).
• Solve 30 questions from old NDA papers on progressions, timed at 60 seconds each.
• Standard sums come back in the binomial theorem; counting ideas continue in permutations and combinations.