Arithmetic Progression (A.P) in Maths ?

An arithmetic progression is a sequence of numbers in which the difference between a term and its preceding term is always constant. This constant difference is called the common difference.

For example, consider the sequence: $2, 5, 8, 11, 14, 17, \dots$

$a_1 = 2,$

$a_2 = 5,$

$a_3 = 8,$

$a_4 = 11$

Now check the differences:

$a2​−a1​=5−2=3$;

$a3​−a2​=8−5=3$;

$a4​−a3​=11−8=3$

Result: Since differences are equal ($3$), it is an A.P.

General Form & $n^{th}$ Term in AP

Standard Notation: $a, a+d, a+2d, a+3d, \dots$

$a$ = first term

$d$ = common difference

Term Formulas:

$1^{st}$ term: $a_1 = a$

$2^{nd}$ term: $a_2 = a+d$

$3^{rd}$ term: $a_3 = a+2d$

General Formula ($n^{th}$ term):$$a_n = a + (n-1)d$$(Applies to any variable: $a_p = a + (p-1)d$ or $a_q = a + (q-1)d$)

3. Sum of Terms ($S_n$)

Formula for sum of first $n$ terms:$$S_n = \frac{n}{2}[2a + (n-1)d]$$

Finding $a_n$ from $S_n$: If the sum is known, the $n^{th}$ term is:$$a_n = S_n – S_{n-1}$$

4. Properties of Three Terms ($a, b, c$)

If $a, b, c$ are in A.P., then common differences must be equal:

$b – a = c – b$

Simplified Relationship:$$2b = a + c$$

Key takeaway: The middle term ($b$) is the average of the first ($a$) and third ($c$) terms.

Basic Simplifications to Solve Arithmetic Progressions

Problem 1: Find the result of subtracting $3$ from $3 + 2\sqrt{2}$.

Solution:

$(3 + 2\sqrt{2}) – 3 = 2\sqrt{2}$

Problem 2: Find the result of subtracting $-4$ from $-8$.

Solution:

$-8 – (-4) = -8 + 4 = -4$

Problem 3: Find the values of $k$ and $k_1$ for the given radical equations.

Solution:

Part A: $\sqrt{8} = k\sqrt{2}$

$\sqrt{4 \times 2} = k\sqrt{2}$

$2\sqrt{2} = k\sqrt{2}$

$k = 2$

Part B: $\sqrt{18} = k_1\sqrt{2}$

$\sqrt{9 \times 2} = k_1\sqrt{2}$

$3\sqrt{2} = k_1\sqrt{2}$

$k_1 = 3$

Problem 4: Solve for $n$: $3n = 81 + 24$

Solution:

$3n = 105$

$n = \frac{105}{3}$

$n = 35$

Problem 5: Find ‘n’ for an Arithmetic Progression where the first term ($a$) is $5$, the common difference ($d$) is $6$, and the last term is $301$.

Solution:

Formula used: $a_n = a + (n-1)d$

$301 = 5 + (n-1)6$

$301 = 5 + 6n – 6$

$301 = 6n – 1$

$301 + 1 = 6n$

$6n = 302$

$n = \frac{302}{6} = \frac{151}{3}$

$n = \frac{151}{3}$

Problem 6: Solve for $n$ from the equation $-18 = (n-1)(-2)$.

Solution:

$(n-1)(-2) = -18$

$(n-1) = \frac{-18}{-2}$

$(n-1) = 9$

$n = 9 + 1$

$n = 10$

Problem 7: Calculate the value of $\frac{100 \times 101}{2}$.

Solution:

$= \frac{10100}{2}$

$= 5050$

Problem 8: Find the sum of an Arithmetic Progression. Given values from previous problems: $n = 21$, $a = 100$, $d = 50$.

Solution:

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

$S = \frac{21}{2} [2 \times 100 + (21-1)50]$

$S = \frac{21}{2} [200 + (20)50]$

$S = \frac{21}{2} [200 + 1000]$

$S = \frac{21}{2} [1200]$

$S = 21 \times 600$

Result: $12600$

Problem 9: Solve the quadratic equation by factoring.

Solution:

Equation: $n^2 – 17n + 52 = 0$

Factor by splitting the middle term (finding factors of $52$ that add to $-17$, which are $-4$ and $-13$):

$n^2 – 4n – 13n + 52 = 0$

$n(n – 4) – 13(n – 4) = 0$

$(n – 4)(n – 13) = 0$

Set each factor to zero:

$n – 4 = 0 \implies n = 4$

$n – 13 = 0 \implies n = 13$

$n = 4$, $n = 13$

Problem 10: Solve the following system of linear equations to find the first term ($a$) and common difference ($d$) of an Arithmetic Progression.

1. $a + 2d = 600$
2. $a + 6d = 700$

Solution:

Using the elimination method, subtract equation (1) from equation (2):

$$=> (a + 6d) – (a + 2d) = 700 – 600$$

$$=> 4d = 100$$

$$=> d = \frac{100}{4}$$

$d = 25$

Next, substitute the value of $d$ back into the first equation to solve for $a$:

$$a + 2(25) = 600$$

$$a + 50 = 600$$

$$a = 600 – 50$$

$a = 550$

The final solution is $a = 550$, $d = 25$.