Arithmetic progression (AP) — nth term, sum of n terms — derivations: Question bank

True or false — justify

Every AP has a positive common difference.

False.

d

can be negative (a decreasing AP like

10, 7, 4, \dots

with

d = - 3

) or zero (a constant sequence

5, 5, 5, \dots

with

d = 0

). The only rule is that

d

is the same every step, not that it is positive.

A sequence with all equal terms (

7, 7, 7, \dots

) is an AP.

True. The difference between consecutive terms is

7 - 7 = 0

, which is constant. A constant

d = 0

still satisfies

a_{n + 1} - a_{n} = d

, so it is a (degenerate) AP.

If three numbers form an AP, the middle one equals the average of the outer two.

True. In an AP

a, b, c

we have

b - a = c - b

, so

2 b = a + c

, giving

b = \frac{a + c}{2}

— exactly the Arithmetic Mean. This is why the middle term is called the arithmetic mean.

Doubling every term of an AP keeps it an AP.

True. If

a_{n} = a + (n - 1) d

, then

2 a_{n} = 2 a + (n - 1) (2 d)

— still linear in

n

, a new AP with first term

2 a

and common difference

2 d

. Any scaling preserves the constant-difference property.

Squaring every term of an AP keeps it an AP.

False. Take

1, 2, 3

(an AP). Squares are

1, 4, 9

, with differences

3

and

5

— not constant. Squaring destroys the linear-in-

n

structure that makes an AP an AP.

S_{n}

of an AP is always a quadratic in

n

Mostly true, with a catch.

S_{n} = \frac{d}{2} n^{2} + (a - \frac{d}{2}) n

has no constant term. If

d = 0

it collapses to the linear

S_{n} = an

. So it is quadratic-with-no-constant-term when

d \neq = 0

, linear when

d = 0

S_{n} = A n^{2} + B n + C

with

C \neq = 0

, the sequence is still an AP.

False. A genuine AP forces

S_{0} = 0

(no terms, no sum), but this formula gives

S_{0} = C \neq = 0

. The nonzero constant term signals the "sequence" is not a true AP from term 1.

For a decreasing AP the sum

S_{n}

eventually decreases as

n

grows.

True. With

d < 0

, later terms turn negative and drag the running total down.

S_{n}

is a downward parabola in

n

, so past its vertex it falls.

The formula

a_{n} = a + (n - 1) d

works for

n = 1

True, and that's the whole point. At

n = 1

it gives

a + (1 - 1) d = a + 0 = a

, the first term. The

(n - 1)

(not

n

) exists precisely so the first term costs zero jumps.

Knowing any two terms of an AP determines the whole AP.

True. Two terms give two equations in the two unknowns

a

and

d

; solving fixes both, and

a, d

rebuild every term. An AP has exactly two degrees of freedom.

Spot the error

A student writes

d = \frac{a _{2}}{a _{1}}

for

2, 5, 8, \dots

Where's the slip?

They divided instead of subtracting. Division gives the ratio of a Geometric Progression (GP). An AP adds a constant, so

d = a_{2} - a_{1} = 3

To find the 20th term of

3, 7, 11, \dots

a student computes

3 + 20 \cdot 4

. What's wrong?

They used

n

jumps instead of

n - 1

. Reaching term 20 takes only

19

steps of

d

, so it should be

3 + 19 \cdot 4 = 79

. Test on term 1:

3 + 0 \cdot 4 = 3

✓, but

3 + 1 \cdot 4

would be wrong.

Someone claims

S_{n} = \frac{n}{2} (a_{1} + a_{n})

only works if the number of terms is even (so they pair up). Correct them.

The formula holds for any

n

, even or odd. The pairing is a derivation aid, not a requirement; algebraically

2 S_{n} = n (a_{1} + a_{n})

regardless of parity — a middle unpaired term just equals the average and fits perfectly.

A student solving "how many terms sum to

84

" gets

n = 6.4

and rounds to

6

. What's the deeper issue?

n

must be a positive integer, so a non-integer means no number of terms gives exactly

84

— you cannot round. The correct conclusion is "no solution," not "approximately 6." (See Quadratic Equations for why both roots must be checked.)

For

S_{n} = 3 n^{2} + 2 n

a student finds

a_{1}

by plugging

n = 1

into

a_{n} = S_{n} - S_{n - 1}

and writes

S_{0}

. Is that valid?

Yes, and it works here because this

S_{n}

has no constant term, so

S_{0} = 0

. In general

a_{1} = S_{1}

directly is the safest first-term rule;

a_{n} = S_{n} - S_{n - 1}

is reliable for

n \geq 2

A student writes the Gauss Summation Trick as: add forwards + backwards, get

n

terms each equal to

a + a_{n}

, so

S_{n} = n (a + a_{n})

. Spot the error.

They forgot to divide by 2. Adding forwards and backwards gives

2 S_{n} = n (a + a_{n})

, so

S_{n} = \frac{n}{2} (a + a_{n})

. The doubled sum must be halved back.

Why questions

Why is the coefficient of

d

always

(n - 1)

and never

n

Because the first term is reached with zero jumps — you start there. Getting from term 1 to term

n

requires

n - 1

equal jumps of size

d

Why does writing

S_{n}

backwards and adding make the sum easy?

As you go forward each term climbs by

d

; the reversed copy drops by

d

at the same spot, so each vertical pair cancels its variation and hits the constant total

a_{1} + a_{n}

. Summing

n

identical values is just multiplication.

Why does "

a_{n}

is linear in

n

" guarantee an AP?

A linear function

a_{n} = p n + q

has constant slope

p

, so

a_{n + 1} - a_{n} = p

is the same every step. Constant consecutive difference is the definition of an AP (see Linear Functions); here

p = d

Why does the average form

S_{n} = n \cdot \frac{a _{1} + a _{n}}{2}

only work for APs and not any sequence?

The average of all terms equals

\frac{first + last}{2}

only when terms are evenly spaced, so highs and lows balance symmetrically. In an uneven sequence the mean is not the midpoint of the extremes.

Why can

S_{n} = \sum_{k = 1}^{n} a_{k}

be turned into a closed formula while a general sum cannot?

The Sigma Notation hides a pattern with constant step

d

; that regularity lets the Gauss pairing collapse

n

additions into one multiplication. A sum with no pattern has no such shortcut.

Why is

a_{n} = S_{n} - S_{n - 1}

true for any sequence, not just APs?

S_{n}

is the total up to term

n

and

S_{n - 1}

the total up to term

n - 1

; their difference is exactly what term

n

contributed. This is a bookkeeping identity, independent of the AP structure.

Edge cases

What is the common difference of a single-term "sequence"

{5}

Undefined — with only one term there is no consecutive pair to subtract. You need at least two terms before

d

has meaning.

d = 0

, what does

S_{n}

become and does the sum formula still hold?

Every term is

a

, so

S_{n} = na

. Plugging

d = 0

into

\frac{n}{2} [2 a + (n - 1) \cdot 0] = \frac{n}{2} \cdot 2 a = na

✓ — the general formula degenerates correctly.

Can an AP's terms cross from positive to negative, and where?

Yes, when

d < 0

. Terms stay positive until

a + (n - 1) d < 0

, i.e. once

n > 1 + \frac{a}{- d}

. The AP is happy to pass through and below zero; nothing special happens to the formulas.

Can

S_{n} = 0

for some

n > 0

even when no single term is zero?

Yes. If positive early terms are exactly cancelled by later negative ones (e.g.

3, 1, - 1, - 3

S_{4} = 0

), the running total returns to zero without any term being zero itself.

For which

n

can

S_{n}

hit its maximum value?

When

d < 0

, at the last

n

before terms go negative — the sum peaks while you're still adding positives. Formally, near where

a_{n} \approx 0

; since

S_{n}

is a downward parabola in

n

, its vertex marks the max.

Is the sequence

1, 1, 2, 3, 5, 8, \dots

(Fibonacci) an AP?

No. Differences are

0, 1, 1, 2, 3, \dots

— not constant. Each term is a sum of the previous two, a different rule entirely, so no fixed

d

exists.

Recall One-line trap summary

Sign of $d$ (any!), $(n - 1)$ not $n$ , divide-by-2 in Gauss, integer- $n$ check, $S_{0} = 0$ test for genuine APs, and "linear $a_{n}$ ⇔ AP." Miss one and the trap closes.

Connections

Parent: AP derivations
Geometric Progression (GP) — the divide-not-subtract confusion.
Arithmetic Mean — the middle-term trap.
Sigma Notation — why closed forms exist.
Quadratic Equations — reject-non-integer- $n$ trap.
Gauss Summation Trick — divide-by-2 trap.
Linear Functions — linear ⇔ AP.