56 Arithmetic Progression

Unlock the Secrets of Arithmetic Progressions!

Unlock the Secrets of Arithmetic Progressions!

Your Friendly Guide to Sequences with a Steady Beat

Hey Future Mathematicians! Ever noticed patterns in numbers? Like counting by 2s (2, 4, 6, 8...) or the seats in a movie theatre increasing by a fixed number in each row? These are examples of a fascinating concept in mathematics called Arithmetic Progression (AP). Let's dive in and explore what they are all about!

What Exactly is an Arithmetic Progression (AP)?

An Arithmetic Progression is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference, usually denoted by 'd'.

Think of it like climbing a staircase where each step is the same height. If you start on a certain step (the first term) and each step up is, say, 15 cm (the common difference), you're moving in an arithmetic progression!

Key Terms to Know:

  • First Term ($a$ or $a_1$): The very first number in the sequence.
  • Common Difference ($d$): The constant amount added to get from one term to the next.
    • If $d > 0$, the AP is increasing. (e.g., 1, 3, 5, 7... where $d=2$)
    • If $d < 0$, the AP is decreasing. (e.g., 10, 7, 4, 1... where $d=-3$)
    • If $d = 0$, all terms are the same. (e.g., 5, 5, 5, 5... where $d=0$)
  • Term: Each number in the sequence.
  • $n$-th Term ($a_n$ or $T_n$): The term at the $n$-th position in the sequence.
Let's see an example: Consider the sequence: 3, 7, 11, 15, 19, ...
  • The first term, $a = 3$.
  • To find the common difference, $d$:
    • $7 - 3 = 4$
    • $11 - 7 = 4$
    • $15 - 11 = 4$
    So, the common difference, $d = 4$.
Since the difference is constant, this is an AP!

Finding the $n$-th Term ($a_n$)

What if you want to find the 100th term of an AP? You wouldn't want to write out all 100 terms! Luckily, there's a formula:

The $n$-th term of an AP is given by:

$$ a_n = a + (n-1)d $$

Where:

  • $a_n$ is the $n$-th term
  • $a$ is the first term
  • $n$ is the term number (position in the sequence)
  • $d$ is the common difference
Example: Find the 10th term of the AP: 2, 5, 8, 11, ...

Solution:

  • First term, $a = 2$.
  • Common difference, $d = 5 - 2 = 3$.
  • We want to find the 10th term, so $n = 10$.

Using the formula $a_n = a + (n-1)d$:

$$ a_{10} = 2 + (10-1) \times 3 $$ $$ a_{10} = 2 + (9) \times 3 $$ $$ a_{10} = 2 + 27 $$ $$ a_{10} = 29 $$

So, the 10th term of this AP is 29.

Sum of the First $n$ Terms of an AP ($S_n$)

Sometimes, we need to find the sum of a certain number of terms in an AP. For instance, if someone saves ₹100 in the first month, ₹150 in the second, ₹200 in the third, and so on (an AP!), how much will they save in a year?

There are two handy formulas for the sum of the first $n$ terms:

Formula 1: When you know the first term ($a$), common difference ($d$), and the number of terms ($n$).

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

Formula 2: When you know the first term ($a$), the last term ($a_n$ or $l$), and the number of terms ($n$).

$$ S_n = \frac{n}{2} [a + a_n] \quad \text{or} \quad S_n = \frac{n}{2} [a + l] $$ (Remember $l$ is just another way to write $a_n$, the last term you're considering).
Example 1 (Using Formula 1): Find the sum of the first 12 terms of the AP: 3, 7, 11, 15, ...

Solution:

  • First term, $a = 3$.
  • Common difference, $d = 7 - 3 = 4$.
  • Number of terms, $n = 12$.

Using $S_n = \frac{n}{2} [2a + (n-1)d]$:

$$ S_{12} = \frac{12}{2} [2(3) + (12-1)4] $$ $$ S_{12} = 6 [6 + (11)4] $$ $$ S_{12} = 6 [6 + 44] $$ $$ S_{12} = 6 [50] $$ $$ S_{12} = 300 $$

So, the sum of the first 12 terms is 300.

Example 2 (Using Formula 2): Find the sum of the AP: 5, 10, 15, ..., 50.

Solution:

  • First term, $a = 5$.
  • Last term, $l = a_n = 50$.
  • Common difference, $d = 10 - 5 = 5$.

First, we need to find $n$ (the number of terms). We use $a_n = a + (n-1)d$:

$$ 50 = 5 + (n-1)5 $$ $$ 50 - 5 = (n-1)5 $$ $$ 45 = (n-1)5 $$ $$ \frac{45}{5} = n-1 $$ $$ 9 = n-1 $$ $$ n = 9 + 1 = 10 $$

So, there are 10 terms.

Now, using $S_n = \frac{n}{2} [a + l]$:

$$ S_{10} = \frac{10}{2} [5 + 50] $$ $$ S_{10} = 5 [55] $$ $$ S_{10} = 275 $$

The sum of this AP is 275.

A Quick Property: Arithmetic Mean

If three numbers $x, y, z$ are in AP, then the middle term $y$ is the arithmetic mean of $x$ and $z$. This means:

$$ y = \frac{x+z}{2} $$

Or, $2y = x+z$. This is because $y-x = d$ and $z-y = d$, so $y-x = z-y$, which simplifies to $2y = x+z$.

Real-Life Rhythms: Where do we see APs?

  • Savings Plans: Saving a fixed extra amount each month.
  • Taxi Fares: A fixed charge for the first kilometer, then a fixed charge for each additional kilometer.
  • Auditorium Seating: Rows of seats where each row has a few more seats than the one in front.
  • Simple Interest Calculations: The total amount year after year if you only earn interest on the principal.

Key Takeaways: Your AP Toolkit!

  • An AP is a sequence with a constant common difference ($d$).
  • To find any term ($a_n$): $a_n = a + (n-1)d$.
  • To find the sum of $n$ terms ($S_n$):
    • $S_n = \frac{n}{2} [2a + (n-1)d]$
    • $S_n = \frac{n}{2} [a + l]$ (if you know the last term $l$)
  • Practice is key! The more problems you solve, the more comfortable you'll become.

Arithmetic Progressions are a fundamental building block in mathematics. Understanding them well will help you in many other areas of study. So, keep practicing, look for patterns, and enjoy the beautiful rhythm of APs!