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3.1 Arithmetic, Geometric & Harmonic Progressions (AP, GP, HP)

Key Takeaways

  • An Arithmetic Progression (AP) has a constant common difference d=an−an−1d = a_n - a_{n-1}, with general term an=a1+(n−1)da_n = a_1 + (n-1)d and partial sum Sn=n2[2a1+(n−1)d]=n2(a1+an)S_n = \frac{n}{2}[2a_1 + (n-1)d] = \frac{n}{2}(a_1 + a_n).

  • A Geometric Progression (GP) has a constant common ratio r=an/an−1r = a_n / a_{n-1}, general term an=a1rn−1a_n = a_1 r^{n-1}, finite sum Sn=a1(1−rn)1−rS_n = \frac{a_1(1-r^n)}{1-r}, and infinite sum S∞=a11−rS_\infty = \frac{a_1}{1-r} valid strictly when ∣r∣<1|r| < 1.

  • A Harmonic Progression (HP) is defined as the sequence of reciprocals of an arithmetic progression; HP terms have no direct closed summation formula and must be analyzed via their reciprocal AP.

  • For any two positive unequal real numbers aa and bb, their arithmetic (AA), geometric (GG), and harmonic (HH) means satisfy A>G>HA > G > H and G2=AHG^2 = AH; if a=ba = b, then A=G=HA = G = H.

  • Standard sigma summation identities ∑k=1nk=n(n+1)2\sum_{k=1}^n k = \frac{n(n+1)}{2}, ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}, and ∑k=1nk3=[n(n+1)2]2\sum_{k=1}^n k^3 = [\frac{n(n+1)}{2}]^2 enable rapid decomposition of complex polynomial series on test day.

Last updated: October 2026

Arithmetic, Geometric & Harmonic Progressions (AP, GP, HP)

Sequences and series form a primary pillar of the mathematics component of the Pakistan Engineering Colleges Admission Test (ECAT). Engineering mathematics demands fluency in identifying structural patterns, evaluating finite and infinite summations, inserting statistical and algebraic means, and decomposing higher-order polynomial series under strict time limits.


1. Mathematical Sequences vs. Series

A sequence is an ordered function f:N→Rf: \mathbb{N} \to \mathbb{R} whose domain is the set of natural numbers. The values f(1),f(2),f(3),…,f(n)f(1), f(2), f(3), \dots, f(n) are denoted as terms a1,a2,a3,…,ana_1, a_2, a_3, \dots, a_n. A series represents the formal indicated sum of the terms of a sequence:

Sn=∑k=1nak=a1+a2+a3+⋯+anS_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \dots + a_n

If the sequence contains a finite count of terms, the series is finite. If the terms continue indefinitely (n→∞n \to \infty), the series is infinite, and its value is governed by the existence of the limit of partial sums lim⁡n→∞Sn\lim_{n \to \infty} S_n.


2. Arithmetic Progressions (AP)

An Arithmetic Progression (AP), or arithmetic sequence, is a sequence in which each term after the first is generated by adding a constant real quantity called the common difference (dd) to the preceding term:

d=ak−ak−1for all k≥2d = a_k - a_{k-1} \quad \text{for all } k \ge 2

General Term (nn-th Term)

Starting with first term a1a_1, successive terms evaluate to a2=a1+da_2 = a_1 + d, a3=a1+2da_3 = a_1 + 2d, and by induction:

an=a1+(n−1)da_n = a_1 + (n-1)d

For any two arbitrary indices mm and nn, the relation generalizes to:

an=am+(n−m)d  ⟹  d=an−amn−ma_n = a_m + (n-m)d \implies d = \frac{a_n - a_m}{n-m}

Sum of the First nn Terms (SnS_n)

The sum of an arithmetic progression can be expressed in two standard equivalent forms:

Sn=n2[2a1+(n−1)d]=n2(a1+an)S_n = \frac{n}{2}\left[2a_1 + (n-1)d\right] = \frac{n}{2}(a_1 + a_n)

Test-Day Shortcut: Quadratic Character of SnS_n

The sum SnS_n of any AP is always a quadratic function in nn with no constant term:

Sn=An2+BnS_n = A n^2 + B n

Expanding the standard summation formula yields:

Sn=(d2)n2+(a1−d2)nS_n = \left(\frac{d}{2}\right)n^2 + \left(a_1 - \frac{d}{2}\right)n

This gives two rapid mental calculation shortcuts:

  1. Common Difference: d=2Ad = 2A (twice the coefficient of n2n^2).
  2. First Term: a1=S1=A+Ba_1 = S_1 = A + B.
  3. nn-th Term Expression: an=Sn−Sn−1=2An+(B−A)a_n = S_n - S_{n-1} = 2An + (B - A).

Example: If Sn=4n2−3nS_n = 4n^2 - 3n, the common difference is instantly d=2(4)=8d = 2(4) = 8, the first term is a1=4−3=1a_1 = 4 - 3 = 1, and an=8n−7a_n = 8n - 7.

Arithmetic Means (AM)

A single Arithmetic Mean (AA) between two numbers aa and bb is the number that places a,A,ba, A, b into an AP:

A−a=b−A  ⟹  A=a+b2A - a = b - A \implies A = \frac{a+b}{2}

To insert nn arithmetic means (A1,A2,…,AnA_1, A_2, \dots, A_n) between aa and bb, the total sequence a,A1,A2,…,An,ba, A_1, A_2, \dots, A_n, b comprises n+2n+2 terms where a1=aa_1 = a and an+2=ba_{n+2} = b:

b=a+(n+2−1)d=a+(n+1)d  ⟹  d=b−an+1b = a + (n+2-1)d = a + (n+1)d \implies d = \frac{b-a}{n+1}

The kk-th inserted mean is:

Ak=a+kd=a+k(b−an+1)A_k = a + k d = a + k\left(\frac{b-a}{n+1}\right)

Sum of nn Inserted Arithmetic Means: The sum of all nn arithmetic means inserted between aa and bb equals nn times the single arithmetic mean between them:

∑k=1nAk=n×(a+b2)=nA\sum_{k=1}^n A_k = n \times \left(\frac{a+b}{2}\right) = n A


3. Geometric Progressions (GP)

A Geometric Progression (GP) is a sequence in which each term after the first is obtained by multiplying the preceding term by a non-zero constant called the common ratio (rr):

r=akak−1for all k≥2,r≠0r = \frac{a_k}{a_{k-1}} \quad \text{for all } k \ge 2, \quad r \ne 0

General Term (nn-th Term)

Given the first term a1a_1 and common ratio rr:

an=a1rn−1a_n = a_1 r^{n-1}

For any two terms at positions mm and nn:

anam=rn−m  ⟹  r=(anam)1n−m\frac{a_n}{a_m} = r^{n-m} \implies r = \left(\frac{a_n}{a_m}\right)^{\frac{1}{n-m}}

Sum of the First nn Terms (SnS_n)

Multiplying Sn=a1+a1r+⋯+a1rn−1S_n = a_1 + a_1 r + \dots + a_1 r^{n-1} by rr and subtracting yields:

Sn=a1(1−rn)1−r=a1(rn−1)r−1(r≠1)S_n = \frac{a_1(1-r^n)}{1-r} = \frac{a_1(r^n-1)}{r-1} \quad (r \ne 1)

When r=1r = 1, Sn=na1S_n = n a_1.

Infinite Geometric Series (S∞S_\infty)

An infinite geometric series converges to a finite sum if and only if ∣r∣<1|r| < 1 (i.e., −1<r<1-1 < r < 1). Under this condition, lim⁡n→∞rn=0\lim_{n \to \infty} r^n = 0, giving:

S∞=lim⁡n→∞a1(1−rn)1−r=a11−r(∣r∣<1)S_\infty = \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r} = \frac{a_1}{1-r} \quad (|r| < 1)

If ∣r∣≥1|r| \ge 1 and a1≠0a_1 \ne 0, the series diverges and no finite sum exists.

Application to Recurring Decimals: To convert 0.18‾=0.181818…0.\overline{18} = 0.181818\dots to a fraction, express it as an infinite GP with a1=18100a_1 = \frac{18}{100} and r=1100r = \frac{1}{100}:

S∞=181001−1100=1810099100=1899=211S_\infty = \frac{\frac{18}{100}}{1 - \frac{1}{100}} = \frac{\frac{18}{100}}{\frac{99}{100}} = \frac{18}{99} = \frac{2}{11}

Geometric Means (GM)

A single Geometric Mean (GG) between two non-zero numbers aa and bb satisfies Ga=bG\frac{G}{a} = \frac{b}{G}, meaning:

G=±abG = \pm \sqrt{ab}

Sign convention: If aa and bb are both positive, G=+abG = +\sqrt{ab}. If aa and bb are both negative, G=−abG = -\sqrt{ab}. If aa and bb have opposite signs, their geometric mean is imaginary.

To insert nn geometric means (G1,G2,…,GnG_1, G_2, \dots, G_n) between aa and bb, the full sequence contains n+2n+2 terms where a1=aa_1 = a and an+2=ba_{n+2} = b:

b=arn+1  ⟹  r=(ba)1n+1b = a r^{n+1} \implies r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}

The kk-th geometric mean is Gk=ark=a(ba)kn+1G_k = a r^k = a \left(\frac{b}{a}\right)^{\frac{k}{n+1}}.

Product of nn Inserted Geometric Means: The product of all nn geometric means inserted between aa and bb equals the nn-th power of their single geometric mean:

∏k=1nGk=G1⋅G2…Gn=(ab)n=Gn\prod_{k=1}^n G_k = G_1 \cdot G_2 \dots G_n = (\sqrt{ab})^n = G^n


4. Harmonic Progressions (HP)

A sequence h1,h2,h3,…,hnh_1, h_2, h_3, \dots, h_n is a Harmonic Progression (HP) if and only if the reciprocals of its terms form an Arithmetic Progression:

1h1,1h2,1h3,…,1hnis an AP\frac{1}{h_1}, \frac{1}{h_2}, \frac{1}{h_3}, \dots, \frac{1}{h_n} \quad \text{is an AP}

General Term (nn-th Term)

Let the reciprocal AP have first term A1=1h1A_1 = \frac{1}{h_1} and common difference d=1h2−1h1d = \frac{1}{h_2} - \frac{1}{h_1}. The nn-th term of the reciprocal AP is:

An=1h1+(n−1)dA_n = \frac{1}{h_1} + (n-1)d

Inverting AnA_n gives the general term of the HP:

hn=1An=11h1+(n−1)d=h1h2h2+(n−1)(h1−h2)h_n = \frac{1}{A_n} = \frac{1}{\frac{1}{h_1} + (n-1)d} = \frac{h_1 h_2}{h_2 + (n-1)(h_1 - h_2)}

Important Examination Note: There is no closed-form algebraic formula for the sum of nn terms of a general harmonic progression (Sn=∑1a+kdS_n = \sum \frac{1}{a + kd}). Never invert the sum formula of an AP to calculate the sum of an HP. Questions on ECAT invariably test term extraction, common differences in the reciprocal domain, or harmonic mean insertion.

Harmonic Mean (HM)

The Harmonic Mean (HH) between two numbers aa and bb is the reciprocal of the arithmetic mean of their reciprocals:

1H=1a+1b2=a+b2ab  ⟹  H=2aba+b\frac{1}{H} = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a+b}{2ab} \implies H = \frac{2ab}{a+b}

To insert nn harmonic means between aa and bb, insert nn arithmetic means between 1a\frac{1}{a} and 1b\frac{1}{b} with common difference d=1b−1an+1=a−b(n+1)abd = \frac{\frac{1}{b} - \frac{1}{a}}{n+1} = \frac{a-b}{(n+1)ab}, and then invert each resulting term.


5. Relationships Between Means (A,G,HA, G, H)

For two positive, distinct real numbers aa and bb (a≠ba \ne b, a,b>0a, b > 0), their three means are:

A=a+b2,G=ab,H=2aba+bA = \frac{a+b}{2}, \quad G = \sqrt{ab}, \quad H = \frac{2ab}{a+b}

1. The Fundamental Identity (G2=AHG^2 = AH)

Multiplying AA and HH:

A⋅H=(a+b2)(2aba+b)=ab=(ab)2=G2A \cdot H = \left(\frac{a+b}{2}\right) \left(\frac{2ab}{a+b}\right) = ab = (\sqrt{ab})^2 = G^2

Because G2=AHG^2 = AH, the geometric mean GG is itself the geometric mean of the arithmetic mean AA and harmonic mean HH. Thus, A,G,HA, G, H form a Geometric Progression in that order.

2. The Inequality of Means (A≥G≥HA \ge G \ge H)

Evaluating the difference A−GA - G:

A−G=a+b2−ab=a+b−2ab2=(a−b)22A - G = \frac{a+b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2}

Because (a−b)2>0(\sqrt{a} - \sqrt{b})^2 > 0 for all real a≠ba \ne b, it follows that A>GA > G.

Next, evaluating the difference G−HG - H:

G−H=ab−2aba+b=ab(a+b)−2aba+b=ab(a+b−2ab)a+b=ab(a−b)2a+bG - H = \sqrt{ab} - \frac{2ab}{a+b} = \frac{\sqrt{ab}(a+b) - 2ab}{a+b} = \frac{\sqrt{ab}(a + b - 2\sqrt{ab})}{a+b} = \frac{\sqrt{ab}(\sqrt{a} - \sqrt{b})^2}{a+b}

Since a,b>0a, b > 0 and a≠ba \ne b, both numerator and denominator are strictly positive, so G>HG > H.

Combining these inequalities:

A>G>H(a≠b,a,b>0)A > G > H \quad (a \ne b, \quad a, b > 0)

If and only if a=ba = b, the equality condition holds: A=G=H=aA = G = H = a.


6. Arithmetico-Geometric Series (AGS)

An Arithmetico-Geometric Series is formed by multiplying corresponding terms of an AP and a GP:

Sn=a+(a+d)r+(a+2d)r2+(a+3d)r3+⋯+[a+(n−1)d]rn−1S_n = a + (a+d)r + (a+2d)r^2 + (a+3d)r^3 + \dots + [a+(n-1)d]r^{n-1}

Sum to Infinity (S∞S_\infty for ∣r∣<1|r| < 1)

To find the infinite sum, write the series and multiply by rr:

S∞=a+(a+d)r+(a+2d)r2+(a+3d)r3+…rS∞=ar+(a+d)r2+(a+2d)r3+…\begin{aligned} S_\infty &= a + (a+d)r + (a+2d)r^2 + (a+3d)r^3 + \dots \\ r S_\infty &= \quad a r + (a+d)r^2 + (a+2d)r^3 + \dots \end{aligned}

Subtracting the second equation from the first:

(1−r)S∞=a+d(r+r2+r3+… )=a+dr1−r(1-r)S_\infty = a + d\left(r + r^2 + r^3 + \dots\right) = a + \frac{dr}{1-r}

Dividing both sides by (1−r)(1-r) gives the standard closed-form formula:

S∞=a1−r+dr(1−r)2(∣r∣<1)S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} \quad (|r| < 1)


7. Sigma (Σ\Sigma) Summation Identities for Natural Numbers

Many entrance examination problems require evaluating sums of polynomials ∑k=1nP(k)\sum_{k=1}^n P(k). These problems rely on the following closed-form identities:

Sum of the First nn Natural Numbers:

∑k=1nk=1+2+3+⋯+n=n(n+1)2\sum_{k=1}^{n} k = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}

Sum of the Squares of the First nn Natural Numbers:

∑k=1nk2=12+22+32+⋯+n2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}

Sum of the Cubes of the First nn Natural Numbers:

∑k=1nk3=13+23+33+⋯+n3=[n(n+1)2]2=(∑k=1nk)2\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + 3^3 + \dots + n^3 = \left[\frac{n(n+1)}{2}\right]^2 = \left(\sum_{k=1}^n k\right)^2

Useful Linear Decompositions:

  • Sum of first nn odd positive integers: ∑k=1n(2k−1)=2[n(n+1)2]−n=n2\sum_{k=1}^n (2k - 1) = 2\left[\frac{n(n+1)}{2}\right] - n = n^2
  • Sum of first nn even positive integers: ∑k=1n2k=2[n(n+1)2]=n(n+1)\sum_{k=1}^n 2k = 2\left[\frac{n(n+1)}{2}\right] = n(n+1)

8. Summary Comparison of Progressions

PropertyArithmetic (AP)Geometric (GP)Harmonic (HP)
Defining Relationak−ak−1=da_k - a_{k-1} = dak/ak−1=ra_k / a_{k-1} = r1/hk−1/hk−1=d1/h_k - 1/h_{k-1} = d
General Term (ana_n)a1+(n−1)da_1 + (n-1)da1rn−1a_1 r^{n-1}11h1+(n−1)d\frac{1}{\frac{1}{h_1} + (n-1)d}
Finite Sum (SnS_n)n2[2a1+(n−1)d]\frac{n}{2}[2a_1 + (n-1)d]a1(1−rn)1−r\frac{a_1(1-r^n)}{1-r}No general closed form
Infinite Sum (S∞S_\infty)Diverges (unless a1=d=0a_1=d=0)a11−r\frac{a_1}{1-r} (if ∣r∣<1\lvert r \rvert < 1)Diverges
Single Mean FormulaA=a+b2A = \frac{a+b}{2}G=±abG = \pm\sqrt{ab}H=2aba+bH = \frac{2ab}{a+b}
Mean PropertyA=1n∑AkA = \frac{1}{n}\sum A_k∏Gk=Gn\prod G_k = G^nH=1AM of reciprocalsH = \frac{1}{\text{AM of reciprocals}}
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Interrelationships Between Progressions and Means
Test Your Knowledge

An infinite geometric series has a first term of 6 and converges to a sum of 18. What is the common ratio rr, and what is the third term a3a_3 of this progression?

A

r=23,a3=83r = \frac{2}{3}, \quad a_3 = \frac{8}{3}

B

r=13,a3=23r = \frac{1}{3}, \quad a_3 = \frac{2}{3}

C

r=12,a3=32r = \frac{1}{2}, \quad a_3 = \frac{3}{2}

D

r=34,a3=278r = \frac{3}{4}, \quad a_3 = \frac{27}{8}

Test Your Knowledge

The Arithmetic Mean (AA) and Harmonic Mean (HH) between two positive distinct numbers are 25 and 16, respectively. What is their Geometric Mean (GG), and what are the two numbers?

A

G=18;numbers are 36 and 14G = 18; \quad \text{numbers are } 36 \text{ and } 14

B

G=20;numbers are 40 and 10G = 20; \quad \text{numbers are } 40 \text{ and } 10

C

G=20;numbers are 30 and 20G = 20; \quad \text{numbers are } 30 \text{ and } 20

D

G=22;numbers are 45 and 5G = 22; \quad \text{numbers are } 45 \text{ and } 5

Test Your Knowledge

The sum of the first nn terms of an arithmetic progression is given by the formula Sn=3n2+5nS_n = 3n^2 + 5n. What are the first term a1a_1 and the common difference dd of this sequence?

A

a1=5,d=3a_1 = 5, \quad d = 3

B

a1=6,d=8a_1 = 6, \quad d = 8

C

a1=8,d=6a_1 = 8, \quad d = 6

D

a1=11,d=6a_1 = 11, \quad d = 6

Sections you finish are checked off in the contents.