Q.150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Arithmetic Progression: The Pattern of Equal Steps

Imagine you're climbing a staircase where every step has the exact same height. If the first step takes you to 3 feet, and each step after that adds exactly 2 feet, your heights would be: 3, 5, 7, 9, 11, ... That's an arithmetic progression — a sequence where you move forward by adding the same number every time.

The Core Idea

An Arithmetic Progression (AP) is a list of numbers where the difference between any two consecutive terms is constant. This constant is called the common difference, usually denoted by $d$.

If the first term is $a$, then the sequence looks like:

The pattern is simple: you start at $a$, then keep adding $d$ to get the next term.

The General Term (nth term)

What if you want the 100th term without writing all 100 numbers? There's a formula.

The first term is $a$ (think of it as $a + 0 \cdot d$).

The second term is $a + d$ (that's $a + 1 \cdot d$).

The third term is $a + 2d$.

Notice the pattern: the term number minus 1 tells you how many times $d$ has been added.

So the nth term (also called the general term) is:

Example: For the AP 3, 5, 7, 9, ... we have $a = 3$, $d = 2$.

The 10th term: $T_{10} = 3 + (10-1) \cdot 2 = 3 + 18 = 21$.

Why "Arithmetic"?

The name comes from an old property: in an AP, every term (except the first and last) is the arithmetic mean of its neighbours. For three consecutive terms $x, y, z$ in an AP:

Check: in 3, 5, 7, we have $5 = \frac{3+7}{2} = 5$. This works for any three consecutive terms.

Sum of the First n Terms

Sometimes you need the total of the first $n$ terms. There's a clever trick.

Write the sum forwards: $S_n = a + (a+d) + (a+2d) + \dots + [a+(n-1)d]$

Write it backwards: $S_n = [a+(n-1)d] + [a+(n-2)d] + \dots + a$

Add them term by term. Each pair adds to $2a + (n-1)d$, and there are $n$ such pairs. So:

Therefore:

There's another useful form. Since the last term $l = a + (n-1)d$, we can write:

This is beautiful: the sum of an AP is just the number of terms times the average of the first and last term.

Example: Sum of first 10 terms of 3, 5, 7, ...

$S_{10} = \frac{10}{2}[2\cdot3 + (10-1)\cdot2] = 5[6 + 18] = 5 \times 24 = 120$

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