Arithmetic Progressions · Class 10 Mathematics: quick revision, 50 bites

What an AP is

Add the same thing every time

An arithmetic progression is a list where each term after the first is the previous term plus a FIXED number.

Takeaway. Same step, every time.

The common difference

Later minus earlier

Takeaway. Reversing it flips the sign.

The general form

Every AP looks like this

Takeaway. The multiplier is always one less than the position.

Check EVERY difference

How to test a list

Takeaway. One bad difference is enough to rule it out.

Two differences are not enough

The shortcut that fails

$3, 7, 11, 14$ starts with 4 and 4 — then the last difference is 3.

Takeaway. Check them all, every time.

The sign of d

Which way round?

For $10, 7, 4, 1$: $d = 7 - 10 = -3$.

Takeaway. Later minus earlier.

Three kinds of AP

What d tells you

Takeaway. All three are genuine APs.

d = 0 is allowed

Yes, it counts

$5, 5, 5, 5$ IS an arithmetic progression, with $d = 0$.

Takeaway. Do not reject a constant list.

2, 4, 8, 16 is not an AP

Adding, not multiplying

Its differences are 2, 4 and 8 — they grow.

Takeaway. A fixed rule is not the same as a fixed difference.

Air in a cylinder

The famous non-example

Takeaway. A quarter of less is less.

d need not be whole

Fractions are fine

$\frac12, 1, \frac32, 2, \ldots$ is an AP with $d = \frac12$.

Takeaway. Nor need a be positive.

From a and d

Building the list

With $a = -5$ and $d = 4$: $-5, -1, 3, 7, \ldots$

Takeaway. Knowing a and d fixes the whole AP.

Finite and infinite

Does it stop?

Takeaway. Topic 3 adds up finite ones.

Arithmetic mean

The middle is the average

Takeaway. For 7, x, 19 in AP, x = 13.

How to answer 'is this an AP?'

Show the differences

Write out every difference, then state the conclusion, then give a and d.

Takeaway. A bare 'yes' earns nothing.

Where APs come from

Older than the textbook

Aryabhata worked with arithmetic progressions more than 1500 years ago, and simple interest still grows as one.

Takeaway. Gauss's schoolboy trick, in Topic 3, uses the same lists.

The nth term

Any term, one line

Takeaway. It reaches the 100th term without writing the other 99.

Why n − 1

The first term is free

You start AT $a_1$, so it costs no steps.

Takeaway. That is the whole reason for the minus one.

a + nd

The error that looks right

Using $a + nd$ gives the term AFTER the one asked for.

Takeaway. Always $(n-1)$, never n.

3, 7, 11, … find the 100th

A hundred terms in one line

Takeaway. 99 steps, not 100.

It subtracts

When d is negative

For $15, 12, 9, \ldots$: $a_{20} = 15 + 19(-3) = 15 - 57 = -42$.

Takeaway. Keep the negative inside the bracket.

Solve for n

Is x a term?

Takeaway. State the reason — that sentence is the mark.

39 yes, 40 no

One step short

In $3, 7, 11, \ldots$: $39$ gives $n = 10$ ✓

Takeaway. Whole and positive, or it is not there.

Put the last term in

Counting a finite AP

Solve $a + (n-1)d = l$ for n.

Takeaway. 29 terms, without listing one.

The +1 trap

One more term than gaps

$\frac{206 - 10}{7} = 28$ counts the STEPS, not the terms.

Takeaway. Use the formula rather than dividing.

Subtract the equations

Two terms given

Takeaway. Substituting into one equation alone goes nowhere.

10th term 41, 18th term 73

A worked pair

$a + 9d = 41$ and $a + 17d = 73$.

Takeaway. The AP is 5, 9, 13, 17, …

The gap between two terms

Distance, not position

Takeaway. The 18th exceeds the 10th by 8d, in every AP.

The nth term from the end

Counting backwards

Takeaway. For 5, 9, …, 81 the 4th from the end is 69.

Answer what was asked

Position or value?

'Which term is 39?' wants a POSITION — the 10th.

Takeaway. Giving the wrong one loses the final mark.

How to set out

Three lines that collect marks

Takeaway. Method marks live in the first two lines.

They fix the whole AP

Two terms are enough

Any two terms determine a and d, and a and d determine every other term.

Takeaway. That is why the subtraction trick always works.

Where it turns up

The same shape as interest

Simple interest and straight-line depreciation are both a starting value plus a fixed number of equal steps.

Takeaway. The nth-term formula computes them directly.

Why the sum collapses

Gauss's trick

Pair the first term with the last, the second with the second-last, and so on.

Takeaway. 50 pairs of 101 gives 5050.

The short form

When the last term is known

Takeaway. Always the quicker choice when l is given.

The long form

When a and d are known

Takeaway. The two forms always agree.

It is (n − 1)d

Inside the bracket too

$S_n = \frac{n}{2}[2a + (n-1)d]$, not $\frac{n}{2}[2a + nd]$.

Takeaway. Nine steps for ten terms.

The natural numbers

Worth memorising

Takeaway. It gives 5050 for n = 100.

First 10 terms of 2, 5, 8, …

A worked sum

Takeaway. Keep n/2 outside until the bracket is done.

Using the short form

Count first, then sum

For $5, 9, \ldots, 81$: first $n = 20$, then $S_{20} = 10(5 + 81) = 860$.

Takeaway. The last term does the work of d.

Solving for n

A given sum makes a quadratic

n appears twice in the formula, so setting $S_n$ equal to a number gives a quadratic in n.

Takeaway. Write the rejection down — it carries a mark.

How many terms sum to 636?

One root survives

For $9, 17, 25, \ldots$: $4n^2 + 5n - 636 = 0$.

Takeaway. A count cannot be negative.

Two valid values of n

Sometimes BOTH are right

With $d < 0$ the later terms are negative and pull the total back down.

Takeaway. Do not discard a positive integer root.

S_n minus S_(n−1)

A term from two sums

Takeaway. And $a_1 = S_1$.

S_n = 3n² + 5n

Straight from the sums

$S_{10} = 350$ and $S_9 = 288$.

Takeaway. Far faster than extracting a and d first.

The question that decides everything

Total or term?

$S_{30}$ is the TOTAL of thirty penalties; $a_{30}$ is the penalty ON day thirty.

Takeaway. Read the question twice before choosing a formula.

Keep n/2 outside

Do not divide early

Finish the bracket first, then multiply by $\frac{n}{2}$.

Takeaway. Fewer fractions, fewer slips.

Choosing between them

Which form do I need?

Takeaway. The short form needs no d.

Total over 30 days

A contractor's penalty

₹200, ₹250, ₹300, … is an AP with $a = 200$, $d = 50$.

Takeaway. The total, not the thirtieth day's ₹1,650.

Where the rule comes from

Older than the story

Aryabhata gave a rule equivalent to this sum formula in the sixth century, long before Gauss's schooldays.

Takeaway. The pairing idea is the same either way.