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Algebra • Sequences and Series

📖 Sequences and Progressions

Sequences and Progressions

Overview

Welcome to the foundational chapter on Sequences and Progressions, a cornerstone of algebra that is absolutely critical for success in the ISI MSQMS entrance examination. This chapter introduces you to ordered lists of numbers that follow specific rules, along with their specialized forms known as progressions. A strong grasp of these concepts is not just about memorizing formulas; it's about developing the analytical skills to identify underlying patterns, predict future terms, and compute sums efficiently.

For aspiring students of ISI, understanding sequences and progressions goes far beyond basic arithmetic. The questions in the entrance exam often require a deep conceptual understanding, the ability to manipulate expressions, and a knack for applying properties in non-standard scenarios. Mastering this chapter will not only secure marks in direct questions but also build a robust analytical framework essential for tackling more advanced topics like limits, series, and even certain aspects of probability and calculus.

Approach this chapter with an emphasis on problem-solving. Focus on understanding the derivation of formulas, the conditions under which they apply, and how to adapt them. The ISI examination frequently tests your ability to quickly recognize the type of progression, apply the most efficient method, and deduce properties from given information. Consistent practice with varied problems is key to transforming theoretical knowledge into high-performance skills.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|---|
| 1 | Introduction to Sequences | Understand sequence definition, patterns, and general terms (ana_n). |
| 2 | Arithmetic Progression (AP) | Analyze constant difference sequences, sums, and terms. |
| 3 | Geometric Progression (GP) | Examine constant ratio sequences, sums, and limits. |
| 4 | Harmonic Progression (HP) | Explore reciprocals of APs and their properties. |

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Learning Objectives

After studying this chapter, you will be able to:

  • Define sequences, identify numerical patterns, and determine general terms (ana_n).

  • Compute the nn-th term and sum of finite Arithmetic and Geometric Progressions.

  • Calculate the sum to infinity for convergent Geometric Progressions.

  • Understand Harmonic Progressions (HP) and solve problems involving relationships between AP, GP, and HP.
  • ---

    Now let's begin with Introduction to Sequences...

    Part 1: Introduction to Sequences

    Key Definitions

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    Essential Formulas

    | Concept | Expression | Use Case |
    |---|---|---|
    | AP n-th Term | an=a1+(n1)da_n = a_1 + (n-1)d | Finding any term in an AP. |
    | AP Sum (n terms) | Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n) or Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d) | Summing terms of an AP. |
    | GP n-th Term | an=a1rn1a_n = a_1 r^{n-1} | Finding any term in a GP. |
    | GP Sum (n terms) | Sn=a1rn1r1S_n = a_1 \frac{r^n - 1}{r - 1} (if r1r \ne 1), Sn=na1S_n = n a_1 (if r=1r = 1) | Summing terms of a GP. |
    | GP Sum (Infinity) | S=a11rS_\infty = \frac{a_1}{1 - r} (if r<1|r| < 1) | Summing an infinite GP. |
    | Sum of First n Naturals | k=1nk=n(n+1)2\sum_{k=1}^n k = \frac{n(n+1)}{2} | Sum of 1+2++n1+2+\dots+n. |
    | Sum of First n Squares | k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} | Sum of 12+22++n21^2+2^2+\dots+n^2. |
    | Sum of First n Cubes | k=1nk3=(n(n+1)2)2\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2 | Sum of 13+23++n31^3+2^3+\dots+n^3. |
    | Relation between SnS_n and ana_n | an=SnSn1a_n = S_n - S_{n-1} for n>1n>1, a1=S1a_1 = S_1 | Finding general term from sum. |
    | Limit of sinxx\frac{\sin x}{x} | limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 | Used in limits involving trigonometry. |
    | Limit of rnr^n | If r>1r>1: \infty, if r=1r=1: 11, if r<1|r|<1: 00, if r1r\le -1: DNE | Determining convergence of geometric sequences. |

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    Must Remember

  • Monotone Convergence Theorem (MCT): If a sequence is monotonically increasing and bounded above, then it converges. If it is monotonically decreasing and bounded below, it also converges. This is crucial for proving convergence without finding the exact limit.

  • Finding General Terms:

  • * Method of Differences: For sequences whose general term is a polynomial in nn, repeated differences will eventually become constant. If the kk-th differences are constant, the general term is a polynomial of degree kk.
    * Mixed Sequences: Some sequences combine polynomial and exponential terms (e.g., an=P(n)+Crna_n = P(n) + C \cdot r^n). Look for patterns in differences that resemble GP terms.
    * Recurrence Relations: For an+1=f(an)a_{n+1} = f(a_n), find fixed points L=f(L)L = f(L) as potential limits. For xn+1=xn1+xnx_{n+1} = \frac{x_n}{1+x_n}, consider 1/xn1/x_n.
  • Cyclicity of Last Digits: The last digits of powers of a number follow a cycle. E.g., for 7n7^n: 71=7,72=499,73=3433,74=24011,7577^1=7, 7^2=49 \to 9, 7^3=343 \to 3, 7^4=2401 \to 1, 7^5 \to 7. The cycle length is 4. To find the last digit of 7k7^k, find k(mod4)k \pmod 4.

  • Telescoping Series: Look for terms that cancel out. Often involves partial fraction decomposition, e.g., 1n(n+1)=1n1n+1\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.

  • Limits of Sums as Integrals (Riemann Sums): limn1nk=1nf(kn)=01f(x)dx\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n f(\frac{k}{n}) = \int_0^1 f(x) dx. Useful for limits of sums like 1n+k\sum \frac{1}{n+k}.

  • Greatest Integer Function (Floor): [x][x] is the largest integer less than or equal to xx. Pay attention to ranges where [x][x] changes value.
  • ---

    Common Mistakes

  • Sequence vs. Series: A sequence is a list of numbers; a series is their sum.

  • Convergence of ana_n vs. Convergence of an\sum a_n: A sequence {an}\{a_n\} can converge (e.g., an=1/na_n = 1/n) while its corresponding series an\sum a_n diverges (e.g., harmonic series).

  • Applying AP/GP formulas blindly: Always verify if the sequence is truly an AP or GP before applying the formulas. Many ISI questions involve non-standard sequences.

  • Incorrectly finding ana_n from SnS_n: Remember an=SnSn1a_n = S_n - S_{n-1} is valid for n>1n>1. a1=S1a_1 = S_1.

  • Ignoring conditions for infinite GP sum: S=a11rS_\infty = \frac{a_1}{1-r} ONLY if r<1|r| < 1.

  • Misinterpreting inequalities for sequences: Ensure you understand the properties of the terms before applying general inequalities.
  • ---

    Quick Practice

    type="MCQ" question="The general term of the sequence 1,4,10,20,35,1, 4, 10, 20, 35, \dots is an=n(n+1)(n+2)ka_n = \frac{n(n+1)(n+2)}{k}. What is the value of kk?" options=["2","3","4","6"] answer="6" hint="Check the first few terms with the general formula and solve for k." solution="Let an=n(n+1)(n+2)ka_n = \frac{n(n+1)(n+2)}{k}.
    For n=1n=1, we have:

    a1=1(2)(3)k=6ka_1 = \frac{1(2)(3)}{k} = \frac{6}{k}

    Given a1=1a_1 = 1, we set them equal:
    6k=1    k=6\frac{6}{k} = 1 \implies k=6

    Let's verify for n=2n=2:
    a2=2(3)(4)6=246=4a_2 = \frac{2(3)(4)}{6} = \frac{24}{6} = 4

    This matches the sequence.
    For n=3n=3:
    a3=3(4)(5)6=606=10a_3 = \frac{3(4)(5)}{6} = \frac{60}{6} = 10

    This also matches.
    The value of kk is 6\boxed{6}.
    "

    type="NAT" question="If Sn=n32n2+5nS_n = n^3 - 2n^2 + 5n is the sum of the first nn terms of a sequence {an}\{a_n\}, find the 5th5^{\text{th}} term a5a_5." answer="48" hint="Use the relation an=SnSn1a_n = S_n - S_{n-1}." solution="We know that an=SnSn1a_n = S_n - S_{n-1}.
    First, calculate S5S_5:

    S5=532(52)+5(5)=1252(25)+25=12550+25=100S_5 = 5^3 - 2(5^2) + 5(5) = 125 - 2(25) + 25 = 125 - 50 + 25 = 100

    Next, calculate S4S_4:
    S4=432(42)+5(4)=642(16)+20=6432+20=52S_4 = 4^3 - 2(4^2) + 5(4) = 64 - 2(16) + 20 = 64 - 32 + 20 = 52

    Now, find a5a_5:
    a5=S5S4=10052=48a_5 = S_5 - S_4 = 100 - 52 = 48

    Alternatively, we can find the general term ana_n first:
    an=SnSn1=(n32n2+5n)((n1)32(n1)2+5(n1))=(n32n2+5n)((n33n2+3n1)2(n22n+1)+(5n5))=(n32n2+5n)(n33n2+3n12n2+4n2+5n5)=(n32n2+5n)(n35n2+12n8)=n32n2+5nn3+5n212n+8=3n27n+8\begin{aligned}a_n & = S_n - S_{n-1} \\
    & = (n^3 - 2n^2 + 5n) - ((n-1)^3 - 2(n-1)^2 + 5(n-1)) \\
    & = (n^3 - 2n^2 + 5n) - ( (n^3 - 3n^2 + 3n - 1) - 2(n^2 - 2n + 1) + (5n - 5) ) \\
    & = (n^3 - 2n^2 + 5n) - ( n^3 - 3n^2 + 3n - 1 - 2n^2 + 4n - 2 + 5n - 5 ) \\
    & = (n^3 - 2n^2 + 5n) - ( n^3 - 5n^2 + 12n - 8 ) \\
    & = n^3 - 2n^2 + 5n - n^3 + 5n^2 - 12n + 8 \\
    & = 3n^2 - 7n + 8\end{aligned}

    Now, substitute n=5n=5 into the formula for ana_n:
    a5=3(52)7(5)+8=3(25)35+8=7535+8=40+8=48a_5 = 3(5^2) - 7(5) + 8 = 3(25) - 35 + 8 = 75 - 35 + 8 = 40 + 8 = 48

    The 5th5^{\text{th}} term a5a_5 is 48\boxed{48}.
    "

    ---

    ubstitute n=5n=5:
    a5=3(52)7(5)+8=3(25)35+8=7535+8=40+8=48a_5 = 3(5^2) - 7(5) + 8 = 3(25) - 35 + 8 = 75 - 35 + 8 = 40 + 8 = 48.
    The original calculation S5=100S_5 = 100 and S4=52S_4 = 52 giving a5=48a_5=48 was correct. The previous check for a1,a2a_1, a_2 was okay. I just got stuck in re-deriving general term ana_n unnecessarily.
    The answer is 48.

    type="MCQ" question="Let x1=3x_1 = 3 and xn+1=12(xn+9xn)x_{n+1} = \frac{1}{2} \left(x_n + \frac{9}{x_n}\right) for n1n \ge 1. What is limnxn\lim_{n \to \infty} x_n?" options=["0","1","3","9"] answer="3" hint="Find the fixed points of the recurrence relation." solution="To find the limit, assume xnLx_n \to L as nn \to \infty. Then

    L=12(L+9L)L = \frac{1}{2} \left(L + \frac{9}{L}\right)

    2L=L+9L2L = L + \frac{9}{L}

    L=9LL = \frac{9}{L}

    L2=9L^2 = 9

    L=±3L = \pm 3

    Since x1=3x_1 = 3 is positive and all subsequent terms xn+1x_{n+1} will also be positive (as it's an average of two positive numbers), the limit must be positive.
    Thus,
    limnxn=3\lim_{n \to \infty} x_n = 3

    This is the Babylonian method for calculating square roots.
    Answer: \boxed{3}"

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    Remember

    > Mastering the art of pattern recognition and understanding convergence criteria (especially the Monotone Convergence Theorem) are paramount for ISI sequence problems.

    See full notes for detailed explanations!

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    Part 2: Arithmetic Progression (AP)

    Key Definitions

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    Essential Formulas

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    Must Remember

  • Linear Relationship: An AP represents a linear function. The nthn^{\text{th}} term an=dn+(a1d)a_n = dn + (a_1-d) shows a linear dependency on nn. This is useful for modeling scenarios with constant rates of change (e.g., depreciation, constant rise).

  • Terms in AP: If a,b,ca, b, c are in AP, the defining property is 2b=a+c2b = a+c. Always use this when three terms are given to be in AP.

  • Odd/Even Indexed Terms: If a1,a2,a3,,ana_1, a_2, a_3, \dots, a_n are in AP with common difference dd, then the terms a1,a3,a5,a_1, a_3, a_5, \dots (odd-indexed terms) also form an AP with common difference 2d2d. Similarly, a2,a4,a6,a_2, a_4, a_6, \dots (even-indexed terms) form an AP with common difference 2d2d.

  • Sum of Absolute Differences: The sum i=1nxai\sum_{i=1}^n |x-a_i| is minimized when xx is the median of a1,a2,,ana_1, a_2, \dots, a_n. If nn is odd, xx is the middle term. If nn is even, xx can be any value between the two middle terms.

  • Logarithm Properties: When AP problems involve logarithms (e.g., logA,logB,logC\log A, \log B, \log C in AP), remember key log rules:

  • * logbM+logbN=logb(MN)\log_b M + \log_b N = \log_b (MN)
    * logbMlogbN=logb(M/N)\log_b M - \log_b N = \log_b (M/N)
    * klogbM=logb(Mk)k \log_b M = \log_b (M^k)
    * Change of base: logbA=logcAlogcb\log_b A = \frac{\log_c A}{\log_c b}
    * Crucial: Arguments of logarithms must always be positive.
  • Symmetry in Summations: For specific functions like f(x)=axax+kf(x) = \frac{a^x}{a^x+k} (where k=ak=a), the property f(x)+f(1x)=1f(x) + f(1-x) = 1 can greatly simplify sums over symmetric intervals, e.g., i=1N1f(iN)\sum_{i=1}^{N-1} f\left(\frac{i}{N}\right). This allows pairing terms.
  • ---

    Common Mistakes

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    Quick Practice

    type="MCQ" question="The 5th5^{\text{th}} term of an AP is 1111 and the 11th11^{\text{th}} term is 2323. What is the 20th20^{\text{th}} term?" options=["39","41","43","45"] answer="41" hint="Use the general term formula to set up a system of equations for a1a_1 and dd." solution="Let the first term be a1a_1 and the common difference be dd.
    Given:

    a5=a1+(51)d=a1+4d=11(1)a_5 = a_1 + (5-1)d = a_1 + 4d = 11 \quad (1)

    a11=a1+(111)d=a1+10d=23(2)a_{11} = a_1 + (11-1)d = a_1 + 10d = 23 \quad (2)

    Subtract (1) from (2):

    (a1+10d)(a1+4d)=2311(a_1 + 10d) - (a_1 + 4d) = 23 - 11

    6d=126d = 12

    d=2d = 2

    Substitute d=2d=2 into (1):

    a1+4(2)=11a_1 + 4(2) = 11

    a1+8=11a_1 + 8 = 11

    a1=3a_1 = 3

    Now find the 20th20^{\text{th}} term:

    a20=a1+(201)d=3+19(2)=3+38=41a_{20} = a_1 + (20-1)d = 3 + 19(2) = 3 + 38 = 41

    The 20th20^{\text{th}} term is 4141.
    Answer: \boxed{41}"

    type="MCQ" question="If log3(x1)\log_3(x-1), log3(x+1)\log_3(x+1), and log3(x+7)\log_3(x+7) are in AP, then xx is:" options=["2","3","4","5"] answer="2" hint="Apply the property of AP for three terms and logarithm rules. Remember to check the domain of logarithms." solution="If log3(x1)\log_3(x-1), log3(x+1)\log_3(x+1), and log3(x+7)\log_3(x+7) are in AP, then the middle term is the arithmetic mean of the other two:

    2log3(x+1)=log3(x1)+log3(x+7)2 \log_3(x+1) = \log_3(x-1) + \log_3(x+7)

    Using logarithm properties:

    log3((x+1)2)=log3((x1)(x+7))\log_3((x+1)^2) = \log_3((x-1)(x+7))

    Since the bases are the same, we can equate the arguments:

    (x+1)2=(x1)(x+7)(x+1)^2 = (x-1)(x+7)

    x2+2x+1=x2+7xx7x^2 + 2x + 1 = x^2 + 7x - x - 7

    x2+2x+1=x2+6x7x^2 + 2x + 1 = x^2 + 6x - 7

    2x+1=6x72x + 1 = 6x - 7

    8=4x8 = 4x

    x=2x = 2

    Check domain for logarithms:
    For x=2x=2:

    x1=21=1>0x-1 = 2-1 = 1 > 0

    x+1=2+1=3>0x+1 = 2+1 = 3 > 0

    x+7=2+7=9>0x+7 = 2+7 = 9 > 0

    All arguments are positive, so x=2x=2 is a valid solution.
    The terms are log31=0\log_3 1 = 0, log33=1\log_3 3 = 1, log39=2\log_3 9 = 2, which are in AP with common difference 1.
    Answer: \boxed{2}"

    type="NAT" question="The sum of the first nn terms of an AP is given by Sn=2n2+3nS_n = 2n^2 + 3n. Find the 10th10^{\text{th}} term of the AP." answer="41" hint="The nthn^{\text{th}} term can be found using the formula an=SnSn1a_n = S_n - S_{n-1}." solution="Given the sum of the first nn terms

    Sn=2n2+3nS_n = 2n^2 + 3n

    To find the nthn^{\text{th}} term,
    an=SnSn1a_n = S_n - S_{n-1}

    First, find S9S_9:
    S9=2(92)+3(9)=2(81)+27=162+27=189S_9 = 2(9^2) + 3(9) = 2(81) + 27 = 162 + 27 = 189

    Now find S10S_{10}:

    S10=2(102)+3(10)=2(100)+30=200+30=230S_{10} = 2(10^2) + 3(10) = 2(100) + 30 = 200 + 30 = 230

    The 10th10^{\text{th}} term a10a_{10} is:

    a10=S10S9=230189=41a_{10} = S_{10} - S_9 = 230 - 189 = 41

    Alternatively, we can find the first term:

    a1=S1=2(1)2+3(1)=2+3=5a_1 = S_1 = 2(1)^2 + 3(1) = 2+3=5

    The second term:
    a2=S2S1=(2(22)+3(2))5=(8+6)5=145=9a_2 = S_2 - S_1 = (2(2^2) + 3(2)) - 5 = (8+6) - 5 = 14-5=9

    The common difference:
    d=a2a1=95=4d = a_2 - a_1 = 9-5=4

    Then the 10th10^{\text{th}} term:
    a10=a1+(101)d=5+9(4)=5+36=41a_{10} = a_1 + (10-1)d = 5 + 9(4) = 5 + 36 = 41

    Answer: \boxed{41}"

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    Remember

    > Arithmetic Progression is fundamentally about linear relationships. Master the general term, sum formulas, and the properties of the arithmetic mean, especially when combined with logarithms or other sequences.

    See full notes for detailed explanations!

    What's Next?

    Review Geometric Progression (GP) and Harmonic Progression (HP), focusing on their definitions, formulas, and interrelationships with AP. Practice problems involving mixed sequences.

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    Part 3: Geometric Progression (GP)

    Key Definitions

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    m after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
    General form: a,ar,ar2,ar3,a, ar, ar^2, ar^3, \dots

    A series formed by multiplying the corresponding terms of an Arithmetic Progression (AP) and a Geometric Progression (GP).
    General form: a,(a+d)r,(a+2d)r2,,[a+(n1)d]rn1a, (a+d)r, (a+2d)r^2, \dots, [a+(n-1)d]r^{n-1}

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    Essential Formulas

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    Must Remember

  • Condition for Infinite Sum Convergence: The sum to infinity S=a1rS_\infty = \frac{a}{1-r} is only valid if and only if the absolute value of the common ratio, r|r|, is strictly less than 1 (i.e., 1<r<1-1 < r < 1). If r1|r| \ge 1, the series diverges, and its sum to infinity does not exist.

  • GP Properties:

  • * If a,b,ca, b, c are in GP, then b2=acb^2 = ac. (Geometric Mean)
    * The product of terms equidistant from the beginning and end is constant: a1an=a2an1=a_1 a_n = a_2 a_{n-1} = \dots.
    * If each term of a GP is raised to the same power, the resulting sequence is also a GP.
    * If a1,a2,a3,a_1, a_2, a_3, \dots are in GP, then loga1,loga2,loga3,\log a_1, \log a_2, \log a_3, \dots are in AP.
  • Arithmetic-Geometric Progression (AGP) Summation: To find the sum of an AGP (SS), multiply the series by its common ratio (rSrS) and subtract rSrS from SS. This transforms the AGP into a standard GP, which can then be summed.

  • * Example: S=a+(a+d)r+(a+2d)r2+S = a + (a+d)r + (a+2d)r^2 + \dots
    * rS=ar+(a+d)r2+(a+2d)r3+rS = ar + (a+d)r^2 + (a+2d)r^3 + \dots
    * SrS=a+dr+dr2+dr3+S - rS = a + dr + dr^2 + dr^3 + \dots (This is aa plus an infinite GP with first term drdr and common ratio rr).
  • Exponential Functions and GP: If a function ff satisfies f(x+y)=f(x)f(y)f(x+y) = f(x)f(y) and f(1)=kf(1)=k, then f(x)=kxf(x)=k^x. For integer xx, this forms a GP: k1,k2,k3,k^1, k^2, k^3, \dots.

  • Special Series Sums:

  • * Sum of nn terms like 1+11+111+1+11+111+\dots: Sn=19k=1n(10k1)=19[10(10n1)9n]S_n = \frac{1}{9} \sum_{k=1}^n (10^k - 1) = \frac{1}{9} \left[ \frac{10(10^n-1)}{9} - n \right].
    * Sum of nn terms like $0.a +

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    Part 4: Harmonic Progression (HP)

    Key Definitions

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    Essential Formulas

    | Formula | Expression | Use Case |
    |---------|---|---|
    | General Term of HP | an=11a1+(n1)da_n = \frac{1}{\frac{1}{a_1} + (n-1)d} where d=1a21a1d = \frac{1}{a_2} - \frac{1}{a_1} is the common difference of the corresponding AP. | To find any term ana_n of an HP. |
    | Condition for a,b,ca,b,c in HP | 2b=1a+1c\frac{2}{b} = \frac{1}{a} + \frac{1}{c} or b=2aca+cb = \frac{2ac}{a+c} | To check if three numbers are in HP, or find the middle term. |
    | Harmonic Mean (HM) of 2 numbers | HM=21a+1b=2aba+bHM = \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b} | To calculate the HM of two numbers. |
    | Harmonic Mean (HM) of nn numbers | HMn=ni=1n1aiHM_n = \frac{n}{\sum_{i=1}^{n} \frac{1}{a_i}} | To calculate the HM of multiple numbers. |
    | Relationship between AM, GM, HM | For positive numbers a1,,ana_1, \ldots, a_n: AMGMHMAM \ge GM \ge HM. For two positive numbers a,ba,b: GM2=AMHMGM^2 = AM \cdot HM. | To establish inequalities or relations between means. |
    | Sum Property of HP | r=1n1arar+1=(n1)a1an\sum_{r=1}^{n-1} a_r a_{r+1} = (n-1)a_1 a_n | To sum products of consecutive terms in an HP (for n2n \ge 2). |

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    Must Remember

  • HP to AP Conversion: The primary strategy for solving any HP problem is to convert it into an AP problem by taking reciprocals. If a1,a2,a_1, a_2, \ldots are in HP, then 1a1,1a2,\frac{1}{a_1}, \frac{1}{a_2}, \ldots are in AP. This is the most crucial step.

  • No Standard Sum Formula: Unlike AP and GP, there is no general formula for the sum of nn terms of an HP. Do not attempt to derive one.

  • AM-GM-HM Inequality: For positive real numbers, AMGMHMAM \ge GM \ge HM. Equality holds if and only if all numbers are equal. For two positive numbers a,ba,b, AM=a+b2AM = \frac{a+b}{2}, GM=abGM = \sqrt{ab}, HM=2aba+bHM = \frac{2ab}{a+b}. Note that GM2=AMHMGM^2 = AM \cdot HM.
  • Recurrence Relations: Be aware that recurrence relations like xn+1=xn1+kxnx_{n+1} = \frac{x_n}{1+kx_n} often imply an HP sequence for xnx_n (by taking reciprocals, 1xn+1=1xn+k\frac{1}{x_{n+1}} = \frac{1}{x_n} + k, which is an AP).

  • Quadratic Equations & HP: If a quadratic equation Ax2+Bx+C=0A x^2 + B x + C = 0 has equal roots and A+B+C=0A+B+C=0, then x=1x=1 is a repeated root. This implies B/(2A)=1-B/(2A) = 1, or B=2AB=-2A. This condition can often simplify to a,b,ca,b,c being in HP if A,B,CA, B, C are expressed in terms of a,b,ca,b,c.

  • Infinite Geometric Series: If x=11ax = \frac{1}{1-a}, y=11by = \frac{1}{1-b}, z=11cz = \frac{1}{1-c} where a,b,ca,b,c are in AP, then x,y,zx,y,z are in HP.
  • ---

    Common Mistakes

    ---

    Quick Practice

    type="MCQ" question="If the pp-th term of an HP is qq and the qq-th term is pp, then its (p+q)(p+q)-th term is:" options=["p+qp+q","pq/(p+q)pq/(p+q)","1/(p+q)1/(p+q)","1"] answer="pq/(p+q)pq/(p+q)" hint="Convert to AP. If AnA_n is the nn-th term of the corresponding AP, then Ap=1/qA_p = 1/q and Aq=1/pA_q = 1/p. Use An=A1+(n1)dA_n = A_1 + (n-1)d to find Ap+qA_{p+q}." solution="Let the HP be a1,a2,a_1, a_2, \ldots. The corresponding AP is A1,A2,A_1, A_2, \ldots where An=1/anA_n = 1/a_n.
    Given ap=qa_p = q, so Ap=1/qA_p = 1/q.
    Given aq=pa_q = p, so Aq=1/pA_q = 1/p.
    Using the formula for AP, An=A1+(n1)dA_n = A_1 + (n-1)d:
    Ap=A1+(p1)d=1/qA_p = A_1 + (p-1)d = 1/q (1)
    Aq=A1+(q1)d=1/pA_q = A_1 + (q-1)d = 1/p (2)
    Subtract (2) from (1):
    (p1)d(q1)d=1/q1/p(p-1)d - (q-1)d = 1/q - 1/p
    (pq)d=pqpq(p-q)d = \frac{p-q}{pq}
    Since pqp \neq q, we can divide by (pq)(p-q):
    d=1pqd = \frac{1}{pq}.
    Substitute dd into (1):
    A1+(p1)1pq=1qA_1 + (p-1)\frac{1}{pq} = \frac{1}{q}
    A1=1qp1pq=p(p1)pq=1pqA_1 = \frac{1}{q} - \frac{p-1}{pq} = \frac{p-(p-1)}{pq} = \frac{1}{pq}.
    Now find Ap+qA_{p+q}:
    Ap+q=A1+(p+q1)d=1pq+(p+q1)1pq=1+p+q1pq=p+qpqA_{p+q} = A_1 + (p+q-1)d = \frac{1}{pq} + (p+q-1)\frac{1}{pq} = \frac{1 + p+q-1}{pq} = \frac{p+q}{pq}.
    The (p+q)(p+q)-th term of HP is ap+q=1/Ap+q=pqp+qa_{p+q} = 1/A_{p+q} = \frac{pq}{p+q}.
    "

    type="NAT" question="The nn-th term of a sequence is ana_n. If a1=1a_1 = 1 and an+1=an1+ana_{n+1} = \frac{a_n}{1+a_n} for n1n \ge 1, then find the 5th term (a5a_5) of the sequence." answer="0.2" hint="Take reciprocals to identify the type of progression." solution="Given a1=1a_1 = 1 and an+1=an1+ana_{n+1} = \frac{a_n}{1+a_n}.
    Take reciprocals of the recurrence relation:
    1an+1=1+anan=1an+1\frac{1}{a_{n+1}} = \frac{1+a_n}{a_n} = \frac{1}{a_n} + 1.
    Let bn=1anb_n = \frac{1}{a_n}. Then bn+1=bn+1b_{n+1} = b_n + 1.
    This shows that the sequence bnb_n is an Arithmetic Progression (AP) with common difference d=1d=1.
    The first term of the AP is b1=1a1=11=1b_1 = \frac{1}{a_1} = \frac{1}{1} = 1.
    The general term of the AP is bn=b1+(n1)d=1+(n1)1=nb_n = b_1 + (n-1)d = 1 + (n-1)1 = n.
    We need to find the 5th term of the original sequence, a5a_5.
    First, find b5b_5:
    b5=5b_5 = 5.
    Since b5=1a5b_5 = \frac{1}{a_5}, we have a5=1b5=15=0.2a_5 = \frac{1}{b_5} = \frac{1}{5} = 0.2.
    "

    ---

    Remember

    > When tackling Harmonic Progression problems, always transform them into Arithmetic Progression problems by taking the reciprocal of each term. This simplifies complex HP scenarios into familiar AP calculations.

    See full notes for detailed explanations!

    ---

    Chapter Summary

    Mastering Sequences and Progressions is foundational for various advanced to

    ---

    pics in ISI. Here are the most critical points to remember:

  • Core Definitions and Formulas: Be fluent with the nthn^{th} term (ana_n) and sum (SnS_n) formulas for Arithmetic Progression (AP) and Geometric Progression (GP). Crucially, understand the condition (r<1|r|<1) and formula (S=a1rS_\infty = \frac{a}{1-r}) for the sum of an infinite GP.

  • Properties are Power: ISI frequently tests understanding of AP, GP, and HP properties.

  • * For AP: If a,b,ca, b, c are in AP, then 2b=a+c2b = a+c.
    * For GP: If a,b,ca, b, c are in GP, then b2=acb^2 = ac.
    * For HP: If a,b,ca, b, c are in HP, then 1a,1b,1c\frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in AP. Always convert HP problems to AP for easier manipulation.
  • Arithmetic, Geometric, and Harmonic Means: Understand the definitions of AM, GM, and HM for two or more numbers. The powerful inequality AMGMHMAM \ge GM \ge HM for positive numbers, along with the relationship GM2=AMHMGM^2 = AM \cdot HM (for two positive numbers), is a frequent tool in optimization and inequality problems. Equality holds if and only if all numbers are equal.

  • Identifying Sequence Types: Practice recognizing whether a given sequence or series is an AP, GP, HP, or can be transformed into one. Look for common differences, common ratios, or reciprocal relationships.

  • Transformations and Recurrence Relations: Some problems involve sequences defined by recurrence relations or require a clever transformation (e.g., taking reciprocals, logarithms, or squaring terms) to reveal an underlying AP or GP.

  • Integrative Problem Solving: ISI questions often combine sequences with other concepts from algebra (e.g., polynomials, quadratic equations), trigonometry, or even calculus. Look for hidden sequence structures within broader mathematical contexts.
  • ---

    Chapter Review Questions

    type="MCQ" question="Let SnS_n denote the sum of the first nn terms of an arithmetic progression. If Smm=Snn\frac{S_m}{m} = \frac{S_n}{n} for distinct positive integers mm and nn, then the common difference dd of the AP must be:" options=["d>0d > 0" "d<0d < 0" "d=0d = 0" "dd can be any real number"] answer="C" hint="Use the formula for SnS_n and simplify the given condition. Remember that mnm \neq n." solution="Let the first term of the AP be aa and the common difference be dd.
    The sum of the first nn terms of an AP is given by Sn=n2(2a+(n1)d)S_n = \frac{n}{2}(2a + (n-1)d).

    Given the condition Smm=Snn\frac{S_m}{m} = \frac{S_n}{n}:

    1m(m2(2a+(m1)d))=1n(n2(2a+(n1)d))\frac{1}{m} \left( \frac{m}{2}(2a + (m-1)d) \right) = \frac{1}{n} \left( \frac{n}{2}(2a + (n-1)d) \right)

    12(2a+(m1)d)=12(2a+(n1)d)\frac{1}{2}(2a + (m-1)d) = \frac{1}{2}(2a + (n-1)d)

    Multiplying by 2 on both sides:
    2a+(m1)d=2a+(n1)d2a + (m-1)d = 2a + (n-1)d

    Subtracting 2a2a from both sides:
    (m1)d=(n1)d(m-1)d = (n-1)d

    Rearranging the terms to one side:
    (m1)d(n1)d=0(m-1)d - (n-1)d = 0

    (m1(n1))d=0(m-1 - (n-1))d = 0

    (m1n+1)d=0(m-1 - n + 1)d = 0

    (mn)d=0(m-n)d = 0

    Since mm and nn are distinct positive integers, mnm \neq n, which implies mn0m-n \neq 0.
    For the product (mn)d(m-n)d to be zero, it must be that d=0d=0.

    Thus, the common difference dd must be 00.

    The final answer is C\boxed{C}"

    type="NAT" question="If a,b,ca, b, c are in Geometric Progression (GP) with common ratio rr, and a,2b,3ca, 2b, 3c are in Arithmetic Progression (AP), find the value of rr given that r1r \neq 1." answer="0.3333333333" hint="Express bb and cc in terms of aa and rr using the GP definition. Then use the AP property 2×(middle term)=sum of other two terms2 \times (\text{middle term}) = \text{sum of other two terms}." solution="Since a,b,ca, b, c are in GP with common ratio rr, we have:

    b=arb = ar

    c=ar2c = ar^2

    Since a,2b,3ca, 2b, 3c are in AP, the property of AP states that 2×(middle term)=sum of other two terms2 \times (\text{middle term}) = \text{sum of other two terms}:

    2(2b)=a+3c2(2b) = a + 3c

    4b=a+3c4b = a + 3c

    Now substitute the expressions for bb and cc from the GP into the AP equation:

    4(ar)=a+3(ar2)4(ar) = a + 3(ar^2)

    Assuming a0a \neq 0 (if a=0a=0, then b=c=0b=c=0, and 0,0,00,0,0 is an AP, which is a trivial case), we can divide the entire equation by aa:

    4r=1+3r24r = 1 + 3r^2

    Rearrange this into a quadratic equation:

    3r24r+1=03r^2 - 4r + 1 = 0

    We can factor this quadratic equation:

    (3r1)(r1)=0(3r-1)(r-1) = 0

    This gives two possible values for rr:

    3r1=0r=133r-1 = 0 \Rightarrow r = \frac{1}{3}

    r1=0r=1r-1 = 0 \Rightarrow r = 1

    The problem states that r1r \neq 1. Therefore, the required value of rr is 13\frac{1}{3}.

    As a decimal, 1/30.33333333331/3 \approx 0.3333333333.
    The final answer is 0.3333333333\boxed{0.3333333333}"

    type="NAT" question="Given three positive real numbers x,y,zx, y, z such that x+y+z=1x+y+z=1. Find the minimum value of 1x+1y+1z\frac{1}{x} + \frac{1}{y} + \frac{1}{z}." answer="9" hint="Recall the relationship between Arithmetic Mean (AM) and Harmonic Mean (HM). For positive numbers a1,a2,,ana_1, a_2, \ldots, a_n, AMHMAM \ge HM." solution="We are given three positive real numbers x,y,zx, y, z such that x+y+z=1x+y+z=1. We need to find the minimum value of 1x+1y+1z\frac{1}{x} + \frac{1}{y} + \frac{1}{z}.

    We can use the AM-HM inequality. For a set of nn positive numbers a1,a2,,ana_1, a_2, \ldots, a_n:

    AM=a1+a2++annAM = \frac{a_1 + a_2 + \ldots + a_n}{n}

    HM=n1a1+1a2++1anHM = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n}}

    The AM-HM inequality states:
    AMHMAM \ge HM

    In our case, the numbers are x,y,zx, y, z. So n=3n=3.

    x+y+z331x+1y+1z\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}

    We are given x+y+z=1x+y+z=1. Substitute this into the inequality:
    1331x+1y+1z\frac{1}{3} \ge \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}

    Now, we want to find the minimum value of 1x+1y+1z\frac{1}{x} + \frac{1}{y} + \frac{1}{z}. Let S=1x+1y+1zS = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}.
    133S\frac{1}{3} \ge \frac{3}{S}

    Since x,y,zx, y, z are positive, SS must also be positive. We can multiply both sides by 3S3S without changing the inequality direction:
    S9S \ge 9

    The minimum value of SS is 99. This minimum occurs when x=y=zx=y=z.
    If x=y=zx=y=z and x+y+z=1x+y+z=1, then 3x=1x=1/33x=1 \Rightarrow x=1/3.
    In this case, 1x+1y+1z=11/3+11/3+11/3=3+3+3=9\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{1/3} + \frac{1}{1/3} + \frac{1}{1/3} = 3+3+3=9.

    The final answer is 9\boxed{9}"

    type="NAT" question="A sequence ana_n is defined by a1=1a_1 = 1 and an+12=an21+an2a_{n+1}^2 = \frac{a_n^2}{1+a_n^2} for n1n \ge 1. Find the value of 1a102\frac{1}{a_{10}^2}." answer="10" hint="Look for a transformation that turns the recurrence relation into a simpler sequence type, such as an AP or GP. Consider taking the reciprocal of an2a_n^2." solution="The sequence is defined by a1=1a_1 = 1 and the recurrence relation an+12=an21+an2a_{n+1}^2 = \frac{a_n^2}{1+a_n^2}.
    We need to find 1a102\frac{1}{a_{10}^2}.

    Let's examine the recurrence relation. It involves an2a_n^2. Let's try to work with 1/an21/a_n^2.
    Take the reciprocal of both sides of the recurrence relation:

    1an+12=1+an2an2\frac{1}{a_{n+1}^2} = \frac{1+a_n^2}{a_n^2}

    Now, split the right-hand side:
    1an+12=1an2+an2an2\frac{1}{a_{n+1}^2} = \frac{1}{a_n^2} + \frac{a_n^2}{a_n^2}

    1an+12=1an2+1\frac{1}{a_{n+1}^2} = \frac{1}{a_n^2} + 1

    Let bn=1an2b_n = \frac{1}{a_n^2}. Then the recurrence relation transforms into:
    bn+1=bn+1b_{n+1} = b_n + 1

    This is the definition of an Arithmetic Progression (AP) with a common difference d=1d=1.

    Now, let's find the first term of this AP, b1b_1:
    Given a1=1a_1 = 1, we have:

    b1=1a12=112=1b_1 = \frac{1}{a_1^2} = \frac{1}{1^2} = 1

    So, we have an AP bnb_n with first term b1=1b_1=1 and common difference d=1d=1.

    The formula for the nthn^{th} term of an AP is bn=b1+(n1)db_n = b_1 + (n-1)d.
    Substituting the values:

    bn=1+(n1)(1)b_n = 1 + (n-1)(1)

    bn=1+n1b_n = 1 + n - 1

    bn=nb_n = n

    We need to find 1a102\frac{1}{a_{10}^2}, which is b10b_{10}.
    Using the formula bn=nb_n=n:
    b10=10b_{10} = 10

    Thus, 1a102=10\frac{1}{a_{10}^2} = 10.

    The final answer is 10\boxed{10}"

    ---

    What's Next?

    Congratulations! You've successfully navigated the fundamental concepts of Sequences and Progressions. This chapter is a cornerstone of quantitative aptitude and forms critical connections across various mathematical domains essential for ISI preparation.

    Key connections to previous and future learning:

    * Foundation in Algebra: This chapter builds directly on your understanding of basic algebraic manipulations, solving equations, and working with functions. A strong grasp here ensures you can handle the algebraic expressions and relations encountered.
    * Calculus (Limits & Series Convergence): The concepts of sequences directly lead into the study of limits of sequences and infinite series. Understanding convergence and divergence of geometric series is a prerequisite for advanced topics like Taylor and Maclaurin series, Fourier series, and other power series expansions.
    * Advanced Algebra & Number Theory: Sequences appear in problems involving roots of polynomials (especially when roots are in AP or GP), solving recurrence relations (common in combinatorics and discrete mathematics), and certain Diophantine equations. The AM-GM-HM inequality is a powerful tool for optimization and proving other inequalities.
    * Problem-Solving & Analytical Skills: Sequences and Progressions train your ability to identify patterns, generalize results, and apply fundamental properties creatively. These skills are invaluable for tackling complex, multi-concept problems often found in the ISI entrance examination.

    Keep practicing and integrating these concepts as you move forward. Your journey through ISI mathematics will frequently revisit these ideas in more sophisticated contexts!

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