Hello, and welcome to today's mathematics lesson. Today, we are diving into Arithmetic Progression. By the end of this lesson, you will understand what makes a sequence arithmetic, how to find any term in such a sequence, and how to calculate the sum of multiple terms.
Let us begin with the foundation. A sequence is simply a group of numbers arranged in a definite order, following a specific rule. The numbers in a sequence are called its terms or elements. For instance, 2, 4, 6, 8, and so on, is a sequence where each number is 2 more than the one before it. Or consider 30, 27, 24, where each term is 3 less than its predecessor. However, 5, 8, 6, 15 has no clear pattern, so it is not a sequence.
When we connect these terms with plus or minus signs, we create a series. So 2 + 4 + 6 + 8 + ... becomes a series. Now, when a sequence follows a uniform rule relating each term to its neighbors, we call it a progression. An arithmetic progression is one such special type.
An Arithmetic Progression, or A.P., is a sequence where every term after the first is obtained by adding a fixed number to the preceding term. Look at 3, 8, 13, 18, 23. Here, each term is 5 more than the previous one. This fixed number is called the common difference, denoted by d. The first term is denoted by a.
Mathematically, if t₁, t₂, t₃, t₄ are consecutive terms, then d = t₂ – t₁ = t₃ – t₂ = t₄ – t₃, and so on. Take 4, 6, 8, 10: here a = 4 and d = 2. Or 4, 8, 12, 16: here d = 4. Conversely, if a = 8 and d = 3, the A.P. becomes 8, 11, 14, 17, and so on.
Notice that the common difference can be negative. In 18, 16, 14, 12, we have d = –2. In –20, –24, –28, we have d = –4. And if d = 0, all terms are identical.
Now, let us discover how to find any term without listing them all. Given first term a and common difference d, the terms are: a, a + d, a + 2d, a + 3d and so on. Observe the pattern: the first term is a + 0d, the second is a + 1d, the third is a + 2d.
Therefore, the nᵗʰ term, denoted tₙ, follows this elegant formula:
tₙ = a + (n – 1)d
This is the general term of an A.P.
Let us apply this. Find the 20ᵗʰ term of the A.P. 9, 5, 1, –3, and so on. Here a = 9 and d = 5 – 9 = –4. So t₂₀ = 9 + (20 – 1)(–4) = 9 – 76 = –67.
Here is another useful idea: if an A.P. has n terms, the last term l equals a + (n – 1)d. Also, the rᵗʰ term from the end equals the (n – r + 1)ᵗʰ term from the beginning. Alternatively, reverse the A.P.: the last term becomes first, and the common difference becomes –d. Then the rᵗʰ term from the end becomes simply the rᵗʰ term of this new A.P.
Now we turn to a powerful result: the sum of the first n terms of an A.P. Denote this sum by Sₙ. Write the sum forward: S = a + (a + d) + (a + 2d) + ... + l. Now write it backward: S = l + (l – d) + (l – 2d) + ... + a.
Add these two equations. Each pair of terms sums to a + l, and there are n such pairs. So 2S = n(a + l), giving us:
S = n/2(a + l)
Since l = a + (n – 1)d, we can also write:
S = n/2[2a + (n – 1)d]
Use the first form when you know the last term, and the second when you do not.
Let us see this in action. Find the sum of the first 10 terms of 8, 4, 0, –4, –8, and so on. Here a = 8, d = –4, and n = 10. So S = 10/2[2(8) + 9(–4)] = 5(16 – 36) = –100.
Or consider: how many terms of 20, 19 1/3, 18 2/3, and so on, must be taken so that their sum is 300? Here a = 20 and d = 19 1/3 – 20 = –2/3. Set up 300 = n/2[40 + (n – 1)(–2/3)]. This simplifies to n² – 61n + 900 = 0, giving n = 25 or 36. Both 25 and 36 are valid answers here.
When dealing with consecutive terms in A.P., clever choices simplify problems. For three consecutive terms with a given sum, use a – d, a, a + d. Their sum is 3a, immediately giving the middle term. For four consecutive terms with a given sum, use a – 3d, a – d, a + d, a + 3d. For five consecutive terms with a given sum: a – 2d, a – d, a, a + d, a + 2d. Notice how the terms are symmetric around the middle, making calculations cleaner.
The arithmetic mean bridges two numbers in an A.P. If a, A, b are in A.P., then A is the arithmetic mean between a and b. From A – a = b – A, we get 2A = a + b, so:
A = (a + b)/2
The arithmetic mean is simply the average of the two numbers.
Finally, let us note two important properties of arithmetic progressions.
Property one: if you add or subtract the same fixed non-zero number from every term of an A.P., the result is still an A.P. Take 5, 8, 11, 14. Add 7 to each: 12, 15, 18, 21 remains an A.P. Subtract 7: –2, 1, 4, 7 is also an A.P.
Property two: if you multiply or divide every term by a given non-zero fixed number, the result is still an A.P. Multiply 5, 8, 11, 14 by 8: 40, 64, 88, 112 is an A.P. Divide 27, 25, 23, 21 by 4: 27/4, 25/4, 23/4, 21/4 remains an A.P. These properties preserve the common difference structure, scaled or shifted.
Let us recap the essential takeaways from today's lesson.
First, an arithmetic progression is a sequence where each term after the first differs from its predecessor by a constant called the common difference d.
Second, the nᵗʰ term is given by tₙ = a + (n – 1)d, where a is the first term.
Third, the sum of the first n terms is Sₙ = n/2(a + l) when the last term l is known, or Sₙ = n/2[2a + (n – 1)d] when it is not.
Fourth, for consecutive terms with a given sum, use symmetric forms like a – d, a, a + d for three terms.
Fifth, the arithmetic mean of two numbers is their average, A = (a + b)/2.
Sixth, adding, subtracting, multiplying, or dividing all terms by a non-zero constant preserves the arithmetic progression structure.
That concludes our exploration of arithmetic progressions. Practice identifying the first term and common difference in any A.P. you encounter. Master the general term formula and the sum formula—they are your powerful tools. Keep questioning, keep calculating, and keep progressing. Until next time, happy learning!