Arithmetic Progressions - Finding the nth term and calculating the sum of first n terms of an Arithmetic Progression

Hello Students! Welcome to this comprehensive guide on Arithmetic Progressions (AP) from your Class 10 NCERT Mathematics syllabus.

Have you ever noticed patterns around you?

• The steps of a ladder get uniformly smaller towards the top.
• Your pocket money increases by a fixed amount every birthday.
• A honeycomb structure has a repeating mathematical order.

In mathematics, when numbers follow a pattern where each term increases or decreases by a fixed value, we call it an Arithmetic Progression. In this chapter, we will master two fundamental skills:

1. Finding any specific term in a sequence (the $n^{\text{th}}$ term).
2. Calculating the total sum of a sequence (the Sum of first $n$ terms).

1. What is an Arithmetic Progression?

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is always constant.

Key Terms to Remember:

• First Term ($a$ or $a_1$): The very first number in the sequence.
• Common Difference ($d$): The constant value added to each term to get the next term. $\text{Common Difference } (d) = a_{k+1} - a_k$
• Number of Terms ($n$): The total count or position of a term in the sequence ($n$ is always a positive integer: $1, 2, 3, \dots$).

💡 Important Note on $d$: The common difference $d$ can be positive (increasing AP), negative (decreasing AP), or zero (constant AP).

Examples:

1. $2, 5, 8, 11, 14, \dots$

   • First term ($a$) = $2$
   • Common difference ($d$) = $5 - 2 = 3$ (Increasing AP)

2. $100, 70, 40, 10, -20, \dots$

   • First term ($a$) = $100$
   • Common difference ($d$) = $70 - 100 = -30$ (Decreasing AP)

3. $5, 5, 5, 5, \dots$

   • First term ($a$) = $5$
   • Common difference ($d$) = $5 - 5 = 0$ (Constant AP)

2. General Form and Finding the $n^{\text{th}}$ Term ($a_n$)

Let's build an AP step-by-step starting with the first term $a$ and adding $d$ repeatedly:

• $1^{\text{st}}$ term ($a_1$): $a = a + (1 - 1)d$
• $2^{\text{nd}}$ term ($a_2$): $a + d = a + (2 - 1)d$
• $3^{\text{rd}}$ term ($a_3$): $a + 2d = a + (3 - 1)d$
• $4^{\text{th}}$ term ($a_4$): $a + 3d = a + (4 - 1)d$

Observing the pattern, notice that the multiplier of $d$ is always one less than the term number!

Formula for the $n^{\text{th}}$ Term:

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

Where:

• $a_n$ = $n^{\text{th}}$ term (also called the general term or last term $l$)
• $a$ = First term
• $n$ = Position of the term
• $d$ = Common difference

3. Sum of the First $n$ Terms ($S_n$)

Imagine your teacher asks you to sum all numbers from $1$ to $100$. Adding them one by one would take forever!

The great mathematician Carl Friedrich Gauss solved this in seconds as a young child by noticing that: $(1 + 100) = 101, \quad (2 + 99) = 101, \quad (3 + 98) = 101 \dots$ There are $50$ such pairs, so the sum is $50 \times 101 = 5050$!

Using this pairing logic, we derive two formulas for finding the sum of the first $n$ terms ($S_n$).

Formula 1: When $a$, $d$, and $n$ are given

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

Formula 2: When the First Term ($a$) and Last Term ($l$) are given

$S_n = \frac{n}{2} (a + l)$ (where $l = a_n = a + (n-1)d$)

4. Useful Secret Formula: Linking $a_n$ and $S_n$

Sometimes questions give you the sum formula $S_n$ in terms of $n$ and ask you to find the $n^{\text{th}}$ term or the AP itself.

Use this quick relation: $a_n = S_n - S_{n-1}$

(The $n^{\text{th}}$ term is equal to the sum of $n$ terms minus the sum of the first $(n-1)$ terms).

Common Pitfalls & Teacher Tips

1. Don't confuse $n$ and $a_n$:
   • $n$ is the position (e.g., $10^{\text{th}}$ step). It must always be a positive whole number ($1, 2, 3\dots$).
   • $a_n$ is the value on that step (e.g., $-15$, $0$, or $45.5$).
2. Watch the Sign of $d$:
   • Always calculate $d$ as $\text{Second term} - \text{First term}$ ($a_2 - a_1$).
   • For decreasing sequences like $20, 16, 12\dots$, $d = 16 - 20 = -4$ (not $+4$).

Practice Questions with Step-by-Step Solutions

Let's test our understanding with 3 board-exam style practice questions!

Question 1 (Finding $a_n$ and checking term existence)

Find the $20^{\text{th}}$ term of the AP: $9, 13, 17, 21, \dots$ Also, check whether $101$ is a term of this AP.

Solution:

Part A: Find the $20^{\text{th}}$ term

1. Identify given values:

   • First term ($a$) = $9$
   • Common difference ($d$) = $13 - 9 = 4$
   • Term position ($n$) = $20$

2. Apply $a_n$ formula: $a_{20} = a + (20 - 1)d$ $a_{20} = 9 + (19 \times 4)$ $a_{20} = 9 + 76 = 85$

Part B: Check if $101$ is a term of the AP

1. Let $a_n = 101$. $101 = a + (n - 1)d$ $101 = 9 + (n - 1)4$ $101 - 9 = (n - 1)4$ $92 = (n - 1)4$ $n - 1 = \frac{92}{4} = 23$ $n = 23 + 1 = 24$

Since $n = 24$ is a positive integer, $101$ is indeed the $24^{\text{th}}$ term of this AP.

Question 2 (Sum of terms when $a_n$ is given)

Find the sum of the first $25$ terms of an AP whose $n^{\text{th}}$ term is given by $a_n = 3 + 4n$.

Solution:

1. Find the first term ($a$) by substituting $n = 1$: $a_1 = 3 + 4(1) = 3 + 4 = 7$ So, $a = 7$.

2. Find the $25^{\text{th}}$ term ($l$) by substituting $n = 25$: $a_{25} = 3 + 4(25) = 3 + 100 = 103$ So, $l = 103$.

3. Use the sum formula $S_n = \frac{n}{2}(a + l)$: $S_{25} = \frac{25}{2} (7 + 103)$ $S_{25} = \frac{25}{2} \times 110$ $S_{25} = 25 \times 55 = 1375$

Final Answer: The sum of the first $25$ terms is $1375$.

Question 3 (Real-World Application Word Problem)

A manufacturer of TV sets produced $600$ sets in the third year and $700$ sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find:

1. The production in the $1^{\text{st}}$ year.
2. The total production in the first $10$ years.

Solution:

Since production increases uniformly, this situation forms an Arithmetic Progression where:

• Year = $n$
• Production in $n^{\text{th}}$ year = $a_n$

1. Form equations from the given information:

   • Production in $3^{\text{rd}}$ year ($a_3$) = $600 \implies a + 2d = 600$ --- (Equation 1)
   • Production in $7^{\text{th}}$ year ($a_7$) = $700 \implies a + 6d = 700$ --- (Equation 2)

2. Solve the linear equations to find $a$ and $d$: Subtract Equation 1 from Equation 2: $(a + 6d) - (a + 2d) = 700 - 600$ $4d = 100 \implies d = 25$

   Substitute $d = 25$ back into Equation 1: $a + 2(25) = 600$ $a + 50 = 600 \implies a = 550$

3. Answer Part 1: Production in the $1^{\text{st}}$ year ($a$) = $550$ sets.

4. Answer Part 2 (Total production in 10 years, $S_{10}$): $S_{10} = \frac{10}{2} \left[ 2(550) + (10 - 1)25 \right]$ $S_{10} = 5 \times \left[ 1100 + 9(25) \right]$ $S_{10} = 5 \times \left[ 1100 + 225 \right]$ $S_{10} = 5 \times 1325 = 6625$

Final Answer:

1. Production in $1^{\text{st}}$ year = $550$ TV sets
2. Total production in first $10$ years = $6,625$ TV sets

Summary Checklist

• $a_n = a + (n - 1)d$ (To find any single term)
• $S_n = \frac{n}{2}[2a + (n - 1)d]$ (To find total sum)
• $S_n = \frac{n}{2}[a + l]$ (Quick sum when last term is known)
• $a_n = S_n - S_{n-1}$ (To find $n^{\text{th}}$ term from sum formula)

Keep practicing questions from your NCERT exercise $5.1$, $5.2$, and $5.3$. You've got this!