A progression is numbers in a definite order by a rule. Spot the rule first: constant difference → AP, constant ratio → GP, reciprocals in AP → HP. The same three means (AM, GM, HM) connect them.
An AP has a constant difference between consecutive terms — the
common difference d.
Example: 2, 5, 8, 11, … — here d = +3.
| Symbol | Meaning |
|---|---|
a |
First term |
d |
Common difference = 2nd − 1st |
n |
Number of terms |
Tn / l
|
n-th / last term |
Sn
|
Sum of first n terms |
Tn = a + (n − 1)d
Sn = (n/2) × [2a + (n − 1)d] — or more usable:
Sn = (n/2) × (first + last)
S = a + (a+d) + … + l and S = l + (l−d) + … + a. Add column-wise:
each column sums to a + l, there are n columns →
2S = n(a+l). This pairing is Gauss's trick — and why averaging first and last
then multiplying by n works.
x, y, z are in AP → 2y = x + z. Here y is the
Arithmetic Mean (AM) of x, z.
(a − d), a, (a + d) — sum is 3a, so a is instantly
known and d cancels. For 4 terms:
(a − 3d), (a − d), (a + d), (a + 3d) (common diff effectively 2d).
(a−d), a, (a+d) → sum 3a = 27 →
a = 9.(9−d)×9×(9+d) = 9(81 − d²) = 648 → 81 − d² = 72 →
d² = 9 → d = ±3.6, 9, 12 (or reversed). The symmetric choice made d cancel in
the sum.
A GP has a constant ratio between consecutive terms — the
common ratio r.
Example: 2, 6, 18, 54, … — here r = 3.
| Symbol | Meaning |
|---|---|
a |
First term |
r |
Common ratio = 2nd / 1st |
Tn = a × rn−1
Sn = a(rn − 1)/(r − 1) if r > 1;
equivalently a(1 − rn)/(1 − r) if r < 1 (same
formula, sign-flipped for positivity).
−1 < r < 1):
S∞ = a / (1 − r) — because rn → 0 as
n → ∞.
|r| ≥ 1, terms don't
shrink — the series diverges (grows without bound or oscillates). Only when each term is a
shrinking fraction does the infinite tail add a finite amount. Example:
1 + 1/2 + 1/4 + 1/8 + … = 2 (here a=1, r=1/2 → S∞=2).
x, y, z are in GP → y² = xz. Here y is the
Geometric Mean (GM).
a³ when using (a/r), a, (ar) — so if
product is given, a is the cube root. For 4 terms:
(a/r³), (a/r), (ar), (ar³) (logically symmetric).
a/r, a, ar → product a³=125 → a=5. If middle term is 5
and sum is e.g. 10.5, then 5/r + 5 + 5r = 10.5 → solve for r. The
symmetric choice again isolates a instantly.
A sequence is in HP if the reciprocals of its terms are in AP.
Example: 1/2, 1/5, 1/8, … is HP because 2, 5, 8, … is AP.
Solving HP problems: no direct sum formula — convert to AP (take reciprocals), solve with AP formulas, then reciprocate the result.
Harmonic Mean (HM): if a and c are two numbers,
their HM b is b = 2ac / (a + c) — which is
2 / (1/a + 1/c), the reciprocal of the average of reciprocals.
2×4×12/(4+12)=96/16=6. Check via AP
reciprocals: 1/4, 1/b, 1/12 in AP → 2/b = 1/4+1/12=1/3 → b=6. Same
result — HM is always between the numbers but closer to the smaller one.
If you have the formula for Sn and need a specific term:
Tn = Sn − Sn−1.
Example: to find the 10th term, compute sum of 10 terms minus sum of 9 terms. This works for any sequence, not just AP/GP — because Sn accumulates all terms up to n.
Sn = 2n² + 3n, find
T5.S5=2×25+15=65, S4=2×16+12=44 →
T5=65−44=21. Also
Tn=Sn−Sn−1=2n²+3n −[2(n−1)²+3(n−1)]=4n−2+3=4n+1
→ T5=21 — matches. Note pattern: coefficient of n² in
Sn determines d (see next).
For series like 1/(1×2) + 1/(2×3) + 1/(3×4) + …
Break via partial fractions:
1/(n(n+1)) = 1/n − 1/(n+1). Most terms cancel (telescope), leaving only first and
last parts.
Sum to n terms = 1 − 1/(n+1) = n/(n+1). For
1/(1×2)+1/(2×3)+1/(3×4)+1/(4×5) =
(1−1/2)+(1/2−1/3)+(1/3−1/4)+(1/4−1/5)=1−1/5=4/5.
If the sum of an AP is given as a quadratic:
Sn = An² + Bn:
d = 2A (always).a = S1 (put n=1). So a = A + B.
Why d = 2A? Because Sn = (d/2)n² + (a − d/2)n from the AP
sum formula. Compare coefficients: A = d/2.
Sn = 3n² + 2n →
d = 6, a = 5. So AP is 5, 11, 17, …. No need to find
Tn first.
| Progression | Condition for x, y, z | Mean |
|---|---|---|
| AP | b − a = c − b → 2b = a + c |
AM = (a + c)/2 |
| GP | b/a = c/b → b² = ac |
GM = √(ac) |
| HP | b = 2ac/(a + c) |
HM = 2ac/(a + c) |
| AM ≥ GM ≥ HM (equality only when a = c) — Arithmetic is largest, Harmonic smallest. | ||
AM-GM-HM intuition: For two positive numbers, AM is the balance point, GM is
the side of the square with same area as rectangle a×c, HM is weighted toward the smaller
number (think average speed for equal distances — harmonic, not arithmetic). Placement
problems love "find GM given AM and HM": use GM² = AM × HM (follows from
definitions).
Three numbers x, y, z in AP → ?
Three numbers x, y, z in GP → ?
Harmonic mean of a and c is?
Sum of 3 AP terms is 30. Fastest choice for terms?
Product of 3 GP terms given — fastest choice?
If Sₙ = 2n² + 3n, common difference d = ?
Telescoping: 1/(1×2)+1/(2×3)+1/(3×4) for n=3 → sum =?
Infinite GP sum exists only when?
Tₙ = Sₙ − Sₙ₋₁ gives?
AP ⇒ constant difference d ; Tn = a+(n−1)d ; Sn = (n/2)(2a+(n−1)d) = (n/2)(first+last)
Property ⇒ x,y,z in AP → 2y=x+z (y is AM)
Selection ⇒ sum of 3 given → (a−d),a,(a+d) ; 4 terms → (a−3d),(a−d),(a+d),(a+3d)
GP ⇒ constant ratio r ; Tn = a·rn−1
Sn ⇒ a(rn−1)/(r−1) (r>1) or a(1−rn)/(1−r) (r<1)
S∞ ⇒ a/(1−r) only when −1<r<1
Property ⇒ x,y,z in GP → y²=xz (y is GM)
Selection ⇒ product of 3 given → (a/r),a,(ar) ; product=a³
HP ⇒ reciprocals in AP (eg 1/2,1/5,1/8)
Solve ⇒ reciprocate → AP → solve → reciprocate back
HM ⇒ b=2ac/(a+c)
Tn ⇒ Sn − Sn−1 (any sequence)
Telescoping ⇒ 1/(n(n+1))=1/n−1/(n+1) → sum=1−1/(n+1)=n/(n+1)
Quadratic Sn=An²+Bn ⇒ d=2A ; a=S₁=A+B
AP: 2b=a+c ; GP: b²=ac ; HP: b=2ac/(a+c)
AM ≥ GM ≥ HM (AM largest, HM smallest)
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