How to Add the Numbers From 1 to 100 in Seconds

Fares that rise by a fixed amount per kilometre, savings that grow by the same sum each month, and theatre rows that gain a fixed number of seats all follow an arithmetic progression. Because the difference never changes, any term and any total can be found without listing.

This part covers the nth term and sum, arithmetic means, symmetric terms, and word problems.

**An AP with first term a and common difference d has nth term $a_n=a+(n-1)d$ and sum $S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)$, where l is the last term.

Deriving the sum.** Write the sum forwards and backwards:

$$S_n=a+(a+d)+\cdots +l,S_n=l+(l-d)+\cdots +a$$

Adding, each of the n pairs sums to $a+l$, so $2S_n=n(a+l)$.

Worked example. For $7,11,15,\dots$ we have $a=7$ and $d=4$:

$$a_{20}=7+19\times 4=83,S_{20}=\frac{20}{2}(7+83)=900$$

Worked example 2. Pairing the first and last numbers, $1+2+\cdots +100=\frac{100}{2}(1+100)=5050$.

An everyday example. An auto-rickshaw fare of ₹30 for the first kilometre and ₹15 for each further kilometre forms an AP, so a 12 km ride costs $30+11\times 15=195$ rupees.

The substance. **$a_n=S_n-S_{n-1}$** — any term can be recovered from a formula for the sum.

**The arithmetic mean of a and b is $\frac{a+b}{2}$, and inserting n arithmetic means between a and b creates an AP of $n+2$ terms with common difference $d=\frac{b-a}{n+1}$.

Single mean.** If a, A, b are in AP, then $A-a=b-A$, so $A=\frac{a+b}{2}$.

Several means. a and b become the first and the $(n+2)$th terms, so $b=a+(n+1)d$.

Worked example. Insert 4 arithmetic means between 3 and 23.

$$d=\frac{23-3}{4+1}=4$$

The means are 7, 11, 15 and 19, giving the AP 3, 7, 11, 15, 19, 23. Their sum, 52, equals 4 times the single mean 13.

An everyday example. Planting 5 saplings evenly between two trees 36 m apart inserts 5 means between 0 and 36, so $d=\frac{36}{6}=6$ m and the saplings stand at 6, 12, 18, 24 and 30 m.

The substance. **Inserting n means creates $n+1$ gaps, not n** — dividing by n is the usual slip.

**Taking three terms as $a-d$, a, $a+d$, or four terms as $a-3d$, $a-d$, $a+d$, $a+3d$, makes the common difference cancel when the terms are added, so the sum gives a immediately.

Symmetric choices:**

- Three terms — $a-d$, a, $a+d$, with sum $3a$
- Four terms — $a-3d$, $a-d$, $a+d$, $a+3d$, with sum $4a$ and common difference $2d$

Worked example. Three numbers in AP have sum 24 and product 440.

- Sum: $3a=24$, so $a=8$
- Product: $(8-d)(8)(8+d)=440$, so $64-d^2=55$ and $d=\pm 3$
- The numbers are 5, 8 and 11

Worked example 2. Four numbers in AP have sum 20 and the sum of their squares is 120. Then $4a=20$ gives $a=5$, and $4a^2+20d^2=120$ gives $d=\pm 1$, so the numbers are 2, 4, 6 and 8.

An everyday example. A right-angled triangular plot whose sides are in AP has sides $a-d$, a, $a+d$; Pythagoras gives $a=4d$, the familiar 3 : 4 : 5 shape.

The substance. With four symmetric terms the common difference is 2d, not d — neighbouring terms are 2d apart.

**A word problem is an AP when a quantity changes by the same amount at each step; identify a, d and n from the situation, then decide whether the question asks for a single term $a_n$ or a running total $S_n$.

An everyday example. Someone saves ₹500 in the first month and increases the saving by ₹100 every month.**

- $a=500$ and $d=100$
- Total in 12 months: $S_{12}=\frac{12}{2}[2(500)+11(100)]=6\times 2100=12600$ rupees
- The month in which ₹1400 is saved: $500+(n-1)100=1400$ gives $n=10$

Worked example. A ladder has rungs 25 cm apart. The rungs shrink uniformly from 45 cm at the bottom to 25 cm at the top, and the top and bottom rungs are 2.5 m apart. There are $\frac{250}{25}+1=11$ rungs, using $\frac{11}{2}(45+25)=385$ cm of wood.

The substance. Check whether the question wants the nth amount or the total so far — confusing $a_n$ with $S_n$ is the commonest word-problem error.