Sequence & Series — NDA Mathematics Formulas and Traps

Arithmetic Progressions — the constant-difference engine

Sequence, series, and the nth term

The two bridges between term and sum

a_n = S_n - S_{n-1}\quad (n \ge 2), \qquad a_1 = S_1

nth term and sum of an AP

nth term and sum

a_n = a + (n-1)d, \qquad S_n = \frac{n}{2}\,[\,2a + (n-1)d\,] = \frac{n}{2}(a + l)

$a$ first term
$d$ common difference
$l$ last term $a_n$

Recovering the term from a sum-formula

Term from sum

a_n = S_n - S_{n-1}\quad (n \ge 2)

The arithmetic mean and symmetric terms

Arithmetic mean of a and b

\text{AM} = \frac{a + b}{2}

The three-term condition and what preserves an AP

Three terms in AP

a,\ b,\ c \text{ in AP} \iff 2b = a + c

Ratio of sums and ratio of terms

Ratio of nth terms from ratio of sums

\frac{a_n}{a_n'} = \frac{f(2n-1)}{g(2n-1)}\quad\text{when}\quad \frac{S_n}{S_n'} = \frac{f(n)}{g(n)}

The clever AP identities

Clever AP identities

S_m = n,\ S_n = m \Rightarrow S_{m+n} = -(m+n) \qquad p\,a_p = q\,a_q \Rightarrow a_{p+q} = 0 \qquad S_p = S_q \Rightarrow S_{p+q} = 0

Common terms of two APs

Common-terms AP

d_{\text{common}} = \operatorname{lcm}(d_1, d_2)

Common traps

The nth term uses $(n-1)d$ , not $nd$

The first term already sits at position 1 with zero steps taken, so reaching position

n

needs

(n-1)

steps:

a_n = a + (n-1)d

. Writing

a + nd

overshoots by one full step. For

2, 5, 8, \ldots

the 10th term is

2 + 9(3) = 29

, not

2 + 10(3) = 32

.

The sum has $(n-1)d$ inside the bracket

The sum is

S_n = \tfrac{n}{2}[\,2a + (n-1)d\,]

— the same

(n-1)

off-by-one lives inside. A frequent slip is

\tfrac{n}{2}[\,2a + nd\,]

. For

3, 7, 11, \ldots

the sum of 20 terms is

\tfrac{20}{2}[6 + 19(4)] = 820

, not

\tfrac{20}{2}[6 + 20(4)] = 860

.

Inserting $k$ means makes $k+1$ gaps, not $k$

Putting

k

arithmetic means between

a

and

b

builds a

(k+2)

-term AP, which has

k+1

equal steps between the endpoints — so the common difference is

d = \dfrac{b-a}{k+1}

, NOT

\dfrac{b-a}{k}

. Inserting 3 means between 2 and 14 gives

d = \tfrac{12}{4} = 3

(means

5, 8, 11

), not

\tfrac{12}{3} = 4

.

Squaring breaks the AP

If

a, b, c

are in AP it does NOT follow that

a^2, b^2, c^2

are in AP, nor that

\tfrac1a, \tfrac1b, \tfrac1c

are. Only adding a constant or scaling by a constant is safe. Test with

1, 2, 3

: the squares

1, 4, 9

are not in AP (

2\cdot4 \ne 1 + 9

).

Geometric Progressions — the constant-ratio engine

Learn this subtopic in the notes

nth term, geometric mean, and the three-term condition

nth term and three-term condition

a_n = a\,r^{\,n-1}, \qquad b^2 = ac \ \text{(for } a,b,c \text{ in GP)}

$a$ first term
$r$ common ratio

Sum of a finite GP

Sum of n terms

S_n = \frac{a\,(r^n - 1)}{r - 1}\quad (r \ne 1)

Sum of an infinite GP

Sum to infinity

S_\infty = \frac{a}{1 - r}\quad (|r| < 1)

Periodic continued fractions & nested radicals

Self-referential equations

x = a + \tfrac{1}{x}\ \Rightarrow\ x^2 - ax - 1 = 0, \qquad x = \sqrt{a + x}\ \Rightarrow\ x^2 - x - a = 0

Product of terms and the middle-term trick

GP product symmetry

a_k \cdot a_{n+1-k} = a_1 \cdot a_n \qquad \prod_{i=1}^{2m-1} a_i = M^{2m-1}

Common traps

The GP nth term is $a\,r^{n-1}$ , not $a\,r^{n}$

The first term has the ratio applied zero times, so position

n

carries

r^{n-1}

:

a_n = a\,r^{\,n-1}

. Using

a\,r^{n}

overshoots by one factor of

r

. For

3, 6, 12, \ldots

the 5th term is

3\cdot 2^{4} = 48

, not

3\cdot 2^{5} = 96

(that is the 6th term).

Repeating-digit sums hide a GP

For

0.3 + 0.33 + 0.333 + \cdots

, write each term as

\tfrac{3}{9}(1 - 10^{-k})

. The sum splits into a constant part (an AP-like count of

\tfrac13

) and a true GP

\sum 10^{-k}

. Don't try to treat the original list as a GP directly — it isn't one.

The convergence condition is not optional

S_\infty = \tfrac{a}{1-r}

is only valid for

|r| < 1

. If a problem's ratio has

|r| \ge 1

, the sum genuinely has no finite value — there is no number to find. Always check

|r|

before reaching for the formula.

It is $\dfrac{a}{1-r}$ , watch the sign in the denominator

The infinite sum is

S_\infty = \dfrac{a}{1-r}

— first term over

(1 - r)

. Flipping it to

\dfrac{a}{r-1}

negates the answer. For

4 + 2 + 1 + \cdots

with

r = \tfrac12

,

S_\infty = \dfrac{4}{1 - \tfrac12} = 8

; the wrong

\dfrac{4}{\tfrac12 - 1} = -8

is negative even though every term is positive.

It is NOT an infinite GP

Reaching for

\tfrac{a}{1-r}

here is wrong — there is no common ratio. The structure is self-referential: set it to

x

, substitute the inner copy, solve the quadratic, keep the positive root.

Harmonic Progressions and the Three Means

Learn this subtopic in the notes

Harmonic progression — flip to the reciprocal AP

HP nth term and three-term condition

a_n = \frac{1}{a + (n-1)d}, \qquad b = \frac{2ac}{a+c}\ \text{(for } a,b,c \text{ in HP)}

AM, GM, HM and the inequality that orders them

The three means and their relation

\text{AM} = \frac{a+b}{2},\quad \text{GM} = \sqrt{ab},\quad \text{HM} = \frac{2ab}{a+b}, \qquad \text{GM}^2 = \text{AM}\cdot\text{HM}

Harmonic mean of several numbers

Harmonic mean of n numbers

\text{HM} = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}

Common traps

You cannot average HP terms directly

The middle term of three numbers in HP is NOT

\tfrac{a+c}{2}

— that is the arithmetic mean. The HP middle is the harmonic mean

b = \dfrac{2ac}{a+c}

(equivalently,

\tfrac1a, \tfrac1b, \tfrac1c

are in AP). For

a = 2, c = 6

the HP middle is

\dfrac{2\cdot 2\cdot 6}{8} = 3

, not

\tfrac{2+6}{2} = 4

. Flip to reciprocals first, always.

AM ≥ GM ≥ HM only for positives

The ordering and the equality-when-equal rule need

a, b > 0

. A common NDA setup gives AM and GM and asks for HM — reach straight for

\text{HM} = \tfrac{\text{GM}^2}{\text{AM}}

rather than solving for

a, b

first.

Don't swap the three mean formulas

Keep them straight:

\text{AM} = \dfrac{a+b}{2}

(sum over 2),

\text{GM} = \sqrt{ab}

(root of the product), and

\text{HM} = \dfrac{2ab}{a+b}

(twice the product over the sum). The classic slip is using

\dfrac{a+b}{2ab}

— that is the reciprocal of the HM, not the HM. For

3

and

6

:

\text{HM} = \dfrac{2\cdot 18}{9} = 4

, not

\dfrac{9}{36} = \tfrac14

.

The identity is $\text{GM}^2 = \text{AM}\cdot\text{HM}$

The geometric mean is the geometric mean of the OTHER two:

\text{GM}^2 = \text{AM}\cdot\text{HM}

, so

\text{GM} = \sqrt{\text{AM}\cdot\text{HM}}

. It is NOT

\text{AM}^2 = \text{GM}\cdot\text{HM}

and the three means are in GP (not AP). Given AM

= 9

, GM

= 6

:

\text{HM} = \dfrac{\text{GM}^2}{\text{AM}} = \dfrac{36}{9} = 4

, not

\dfrac{\text{AM}^2}{\text{GM}} = \dfrac{81}{6}

.

Interrelating AP, GP and HP — the bridge tricks

Learn this subtopic in the notes

The three three-term conditions

The unifying ratio

\frac{a-b}{b-c} = \begin{cases} 1 & \text{AP} \\[2pt] a/b & \text{GP} \\[2pt] a/c & \text{HP} \end{cases}

The log bridge: a GP becomes an AP

The bridge

x, y, z \text{ in GP} \iff \log x, \log y, \log z \text{ in AP}

The reciprocal bridge: an HP becomes an AP

Reciprocal flip

\frac{1}{u}, \frac{1}{v}, \frac{1}{w} \text{ in HP} \iff u, v, w \text{ in AP}

Roots, coefficients, and progression conditions

Vieta's relations (monic-friendly)

\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}

Special Series and Special Sums

Learn this subtopic in the notes

Sums of powers of natural numbers

The three power sums

\sum k = \frac{n(n+1)}{2},\quad \sum k^2 = \frac{n(n+1)(2n+1)}{6},\quad \sum k^3 = \left[\frac{n(n+1)}{2}\right]^2

Factorial sums — telescoping and remainders

Factorial telescoping

n\cdot n! = (n+1)! - n! \;\Rightarrow\; \sum_{k=1}^{n} k\cdot k! = (n+1)! - 1

Telescoping sums, repunits, and divisibility patterns

Telescoping standard sum

\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}

Common traps

Don't confuse the three power-sum formulas

Keep them distinct:

\sum k = \dfrac{n(n+1)}{2}

,

\sum k^2 = \dfrac{n(n+1)(2n+1)}{6}

, and

\sum k^3 = \left[\dfrac{n(n+1)}{2}\right]^2

. Only the cube-sum is a square. A frequent error is using the

\sum k

formula where

\sum k^2

is needed — the squares formula carries the extra

(2n+1)

and divides by 6. For

n = 4

:

\sum k^2 = \dfrac{4\cdot 5\cdot 9}{6} = 30

, not

\dfrac{4\cdot 5}{2} = 10

.

Sequence & Series formulas

Arithmetic Progressions — the constant-difference engine

Sequence, series, and the nth term

nth term and sum of an AP

Recovering the term from a sum-formula

The arithmetic mean and symmetric terms

The three-term condition and what preserves an AP

Ratio of sums and ratio of terms

The clever AP identities

Common terms of two APs

Common traps

Geometric Progressions — the constant-ratio engine

nth term, geometric mean, and the three-term condition

Sum of a finite GP

Sum of an infinite GP

Periodic continued fractions & nested radicals

Product of terms and the middle-term trick

Common traps

Harmonic Progressions and the Three Means

Harmonic progression — flip to the reciprocal AP

AM, GM, HM and the inequality that orders them

Harmonic mean of several numbers

Common traps

Interrelating AP, GP and HP — the bridge tricks

The three three-term conditions

The log bridge: a GP becomes an AP

The reciprocal bridge: an HP becomes an AP

Roots, coefficients, and progression conditions

Special Series and Special Sums

Sums of powers of natural numbers

Factorial sums — telescoping and remainders

Telescoping sums, repunits, and divisibility patterns

Common traps

More NDA Mathematics formula sheets