CBSE Class 10 Mathematics Arithmetic Progression Notes Set 05

Sequences and Arithmetic Progression

We have come across various number patterns in our earlier classes. For example: 1, 2, 3, 4, ... (each term is 1 more than the previous one); −3, −2, −1, 0, ... (each term is obtained by adding 1); 100, 70, 40, 10, ... (each term is 30 less than the previous one); 2, 2, 2, 2, ... (each term is obtained by adding 0); −1.0, −1.5, −2.0, −2.5, ... (each term is obtained by adding −0.5).

In all these lists the terms are arranged in some definite order. Such patterns are called sequences. Each of the numbers in a sequence is called a term of the sequence and is denoted by a₁, a₂, a₃, ... etc.

Example 1. Write the first three terms of the sequences defined by (a) aₙ = 3n + 2 and (b) aₙ = n² + 1.
Answer:
(a) Putting n = 1, 2, 3: a₁ = 3 × 1 + 2 = 5, a₂ = 3 × 2 + 2 = 8, a₃ = 3 × 3 + 2 = 11. The first three terms are 5, 8 and 11.
(b) a₁ = 1² + 1 = 2, a₂ = 2² + 1 = 5, a₃ = 3² + 1 = 10. The first three terms are 2, 5 and 10.

Arithmetic Progression (A.P.)

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is known as the common difference.

Let us denote the first term of an A.P. by a₁, the second term by a₂, ..., the nth term by aₙ and the common difference by d. Then the A.P. is a₁, a₂, a₃, ..., aₙ and

\( a_2-a_1=a_3-a_2=\dots=a_n-a_{n-1}=d \).

General form of an A.P.: a, a + d, a + 2d, a + 3d, ... Here a is the first term and d is the common difference.

Note:
1. The number of terms n in an A.P. always belongs to the set of natural numbers.
2. The common difference d can be zero, positive or negative.
3. If d = 0, all terms of the A.P. are the same, e.g. 3, 3, 3, 3, ... (d = 0).
4. If d is positive, the A.P. is increasing, e.g. 6, 9, 12, 15, ... (d = +3).
5. If d is negative, the A.P. is decreasing, e.g. 100, 70, 40, 10, ... (d = −30).

Knowledge Hub
1. In an A.P., if a constant is added to or subtracted from each term, the resulting sequence is also an A.P. with the same common difference.
2. In an A.P., if each term is multiplied or divided by a non-zero constant k, the resulting sequence is also an A.P. with common difference kd, where d is the common difference of the given A.P.
3. If three terms a, b and c are in A.P., then 2b = a + c.

Finite and Infinite A.P.

1. Finite A.P.: An A.P. in which the number of terms is finite, e.g. 2, 4, 6, 8, 10, 12, 14.
2. Infinite A.P.: An A.P. in which the number of terms is infinite, e.g. 5, 10, 15, 20, ...

nth Term of an A.P.

Let a₁, a₂, a₃, ... be the terms of an A.P. whose first term is a₁ = a and common difference is d. Then the second term a₂ = a₁ + d, the third term a₃ = a₂ + d, the fourth term a₄ = a₃ + d, and so on. Hence the nth term of the A.P. with first term a and common difference d is

\[ a_n=a+(n-1)d \]

Note: aₙ is also known as the general term of the A.P. The last term of a finite A.P. is denoted by l and l = a + (n − 1)d.

mth Term of an A.P. from the End

We can find the mth term of an A.P. from the end by taking the last term (if given) as the first term and the negative of the common difference as the common difference, that is, the reverse A.P. Hence

\[ m\text{th term from the end}=l+(m-1)(-d)=l-(m-1)d \]Also, the (n − m + 1)th term from the beginning is the mth term from the end of an A.P. with n terms.

Terms Selection in an A.P.

It is convenient to select the terms as per the following table:

Example 2. The nth term of a sequence is 3n − 2. Is this sequence an A.P.? If so, find its 10th term.
Answer:
Putting n = 1, 2, 3: a₁ = 1, a₂ = 4, a₃ = 7. Then a₂ − a₁ = 3 = a₃ − a₂, and in general aₙ₊₁ − aₙ = 3(n + 1) − 2 − (3n − 2) = 3, a constant.
Hence the sequence is an A.P. with d = 3. a₁₀ = 3 × 10 − 2 = 28.

Example 3. If an A.P. is given by 9, 13, 17, 21, 25, ..., find the 12th term, the 24th term and the nth term.
Answer:
Here a = 9 and d = 4.
a₁₂ = 9 + 11 × 4 = 53 and a₂₄ = 9 + 23 × 4 = 101.
aₙ = 9 + (n − 1) × 4 = 4n + 5.

Example 4. Which term of the sequence 4, 9, 14, 19, ... is 124?
Answer:
a = 4, d = 5. Let aₙ = 124: 4 + (n − 1) × 5 = 124 ⇒ n − 1 = 24 ⇒ n = 25.
Therefore 124 is the 25th term of the sequence.

Example 5. If five times the fifth term of an A.P. is equal to 8 times its eighth term, show that its 13th term is zero.
Answer:
5a₅ = 8a₈ ⇒ 5(a + 4d) = 8(a + 7d) ⇒ 5a + 20d = 8a + 56d ⇒ 3a + 36d = 0 ⇒ a + 12d = 0.
a₁₃ = a + 12d = 0. Hence proved.

Example 6. Find the 10th term from the end of the A.P. 4, 9, 14, ..., 254.
Answer:
Here l = 254 and d = 5. The 10th term from the end = l − (10 − 1)d = 254 − 9 × 5 = 209.