Introduction to Sequences and Arithmetic Progression (AP)

What are Sequences and Series?

• Sequence: A sequence is an arrangement of numbers in a definite order, according to some rule. Formally, a sequence is a function whose domain is the set of natural numbers. We denote the terms by $a_1, a_2, a_3, ..., a_n$.
• Series: A series is the sum of the terms of a sequence. It is represented by the expression $a_1 + a_2 + a_3 + ... + a_n$.

Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence in which the difference between any two consecutive terms is constant. This constant is called the common difference (d).

• Example: The sequence 2, 5, 8, 11, … is an AP with a first term $a=2$ and a common difference $d=3$.

Key Formulas for AP:

• General Term (n-th term): The n-th term of an AP is given by:

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

where 'a' is the first term and 'd' is the common difference.

• Sum of first n terms: The sum of the first n terms of an AP is given by:

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

Alternatively, if 'l' is the last term ($a_n$):

$$S_n = \frac{n}{2}[a + l]$$

Properties of AP:

• If a constant is added to, subtracted from, multiplied by, or divided by each term of an AP, the resulting sequence is also an AP.

• The sum of terms equidistant from the beginning and end is constant and equal to the sum of the first and last terms ($a_1 + a_n = a_2 + a_{n-1}$, etc.).

• Selection of Terms: For problems involving sums or products, it is convenient to select terms as:

  • 3 terms: $a-d, a, a+d$
  • 4 terms: $a-3d, a-d, a+d, a+3d$
  • 5 terms: $a-2d, a-d, a, a+d, a+2d$

• Arithmetic Mean (AM): The AM of two numbers a and b is $\frac{a+b}{2}$. Inserting 'n' arithmetic means between a and b means creating an AP of n+2 terms where 'a' is the first term and 'b' is the last.

Example 1: Finding a Specific Term

Question: Find the 10th term of the AP: 2, 7, 12, …

Solution:

Here, the first term $a=2$ and the common difference $d = 7-2=5$. We need to find the 10th term, so n=10.

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

$$a_{10} = 2 + (10-1)5 = 2 + 9 \times 5 = 2 + 45 = 47$$

Example 2: Sum of an AP

Question: Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5.

Solution:

The required numbers are 105, 110, 115, …, 995. This is an AP with first term $a=105$, common difference $d=5$, and last term $l=995$.

First, find the number of terms (n). $a_n = a + (n-1)d \implies 995 = 105 + (n-1)5 \implies 890 = (n-1)5 \implies n-1 = 178 \implies n=179$.

Now, find the sum: $S_n = \frac{n}{2}(a+l) = \frac{179}{2}(105+995) = \frac{179}{2}(1100) = 179 \times 550 = 98450$.

Example 3: Classic AP Property Problem

Question: If the p-th term of an AP is q and the q-th term is p, find the (p+q)-th term.

Solution:

Given: $a_p = a + (p-1)d = q$ …(1)

And $a_q = a + (q-1)d = p$ …(2)

Subtracting (2) from (1): $(p-1)d - (q-1)d = q-p \implies (p-q)d = -(p-q)$. Since $p \ne q$, we get $d=-1$.

Substitute $d=-1$ in (1): $a + (p-1)(-1) = q \implies a - p + 1 = q \implies a = p+q-1$.

Now, find the (p+q)-th term:

$a_{p+q} = a + (p+q-1)d = (p+q-1) + (p+q-1)(-1) = (p+q-1) - (p+q-1) = 0$.

Example 4: Selection of Terms

Question: The sum of three numbers in an AP is -3, and their product is 8. Find the numbers.

Solution:

Let the three numbers be $a-d, a, a+d$.

Sum: $(a-d) + a + (a+d) = -3 \implies 3a = -3 \implies a = -1$.

Product: $(a-d)(a)(a+d) = 8$. Substitute $a=-1$: $(-1-d)(-1)(-1+d) = 8$.

$(-1)(d+1)(d-1) = 8 \implies -(d^2-1) = 8 \implies -d^2+1=8 \implies d^2=-7 \implies d=\pm i\sqrt{7}$. The numbers are not real. Let's assume the product is -8. Then $-(d^2-1)=-8 \implies d^2-1=8 \implies d^2=9 \implies d=\pm 3$.

If $d=3$, the numbers are -4, -1, 2. If $d=-3$, the numbers are 2, -1, -4. The numbers are -4, -1, 2.