Arithmetic Progressions Class 10 Notes: nth Term and Sum
These arithmetic progressions class 10 notes cover Chapter 5 for the current academic session. An arithmetic progression (AP) is a list of numbers where each term increases or decreases by a fixed amount. This chapter builds two formulas: one for the nth term and one for the sum of n terms.
What Is a Pattern? From Daily Life to Arithmetic Progressions
The chapter starts by noticing patterns in nature like honeycomb holes, sunflower petals, and spirals on a pineapple. It then looks at patterns from daily life. Some patterns are made by adding a fixed number, some by multiplying by a fixed number, and some by squaring consecutive numbers like \(1^2, 2^2, 3^2, \dots\) (NCERT, p. 1).
Not all number patterns in daily life are APs. For example, the sequence \(1, 1, 2, 3, 5, 8, \dots\) is a famous pattern where each term is the sum of the previous two. This is not an AP because the difference between terms keeps changing.
Another savings scheme example shows the amounts 10000, 12500, 15625, 19531.25. This pattern grows by multiplying by \(\frac{5}{4}\) each time, which means it is not an AP.
The squares pattern is shown in the figure below. The squares have side lengths 1, 2, 3, and 4 units, containing 1, 4, 9, and 16 unit squares. The difference between the number of squares is not constant, so this is not an AP either.
Arithmetic progressions study only the simplest pattern: one where you add a fixed number to get the next term. As you review, check our broader Class 10 Mathematics notes and the CBSE notes hub for related chapters.
Arithmetic Progressions Class 10 Notes: Basic Definition
Arithmetic Progression (AP) is a list of numbers where each term is obtained by adding a fixed number to the preceding term (NCERT, p. 3). The first term is called a, and the fixed number added each time is the common difference (d). You cannot write an AP by knowing only the first term or only the common difference; you need both.
| Term | Meaning | Example |
|---|---|---|
| Term | Each number in the list of an AP. | In 4, 7, 10, 13, the term \(a_3 = 10\) |
| Common difference (d) | The fixed number added to a term to get the next term. Found by \(a_{k+1} – a_k\). Can be positive, negative, or zero. | For 4, 7, 10, 13, \(d = 7 – 4 = 3\) |
| General form | The standard way to write an AP using the first term \(a\) and common difference \(d\): \(a, a+d, a+2d, a+3d, \dots\) | If \(a=6, d=2\), general form is 6, 8, 10, 12… |
| Finite AP | An AP that has a last term. | 2, 5, 8, 11 (ends at 11) |
| Infinite AP | An AP that continues forever without a last term. | 1, 2, 3, 4, … |
An important result from this chapter is that \(a_n = S_n – S_{n-1}\) (NCERT, p. 24). This means the nth term of an AP is equal to the sum of the first n terms minus the sum of the first (n-1) terms. This is a useful identity for solving questions where you are given a formula for \(S_n\) in terms of n.
How to Check Whether a Given List Is an AP
To test whether a list of numbers forms an AP, subtract the kth term from the (k+1)th term. If the difference \(a_{k+1} – a_k\) is the same constant for every pair of consecutive terms, the list is an AP. For a decreasing sequence, always subtract the earlier term from the later term, even if the result is negative (NCERT, p. 4).
Example: Verify the list 7, 9.5, 12, 14.5 is an AP
Step 1: Find \(a_2 – a_1\) by subtracting the first term from the second.
\[ a_2 – a_1 = 9.5 – 7 = 2.5 \]
Step 2: Find \(a_3 – a_2\).
\[ a_3 – a_2 = 12 – 9.5 = 2.5 \]
Step 3: Find \(a_4 – a_3\).
\[ a_4 – a_3 = 14.5 – 12 = 2.5 \]
Final answer: Since the difference is the same constant (2.5) each time, the list is an AP with \(a = 7\) and \(d = 2.5\).
The nth Term Formula and How to Use It
By adding \(d\) repeatedly to the first term, we see a pattern: \(a_2 = a + 1d\), \(a_3 = a + 2d\), \(a_4 = a + 3d\), and generally, the nth term is \(a_n = a + (n-1)d\) (NCERT, p. 8). Here, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the term number. \(n\) must be a positive integer.
Example 1: Find the 14th term of an AP
Method: Use the nth term formula \(a_n = a + (n-1)d\).
Step 1: Substitute \(a = -3\), \(d = 4\), and \(n = 14\).
\[ a_{14} = -3 + (14 – 1) \times 4 \]
Step 2: Simplify the bracket.
\[ a_{14} = -3 + 13 \times 4 = -3 + 52 \]
Final answer: The 14th term is 49.
Example 2: Which term of 50, 45, 40, … is 5?
Method: Equate the nth term formula to 5 and solve for n.
Step 1: Identify \(a = 50\), \(d = 45 – 50 = -5\), and \(a_n = 5\).
\[ 5 = 50 + (n – 1)(-5) \]
Step 2: Solve the equation for n.
\[ -45 = (n – 1)(-5) \]
\[ 9 = n – 1 \implies n = 10 \]
Final answer: The number 5 is the 10th term of the AP. Because n is a positive integer, 5 is a valid term.
Finding a Term From the End of an AP
To find the kth term from the end, you can reverse the AP. The last term becomes the first term, and the common difference changes sign (NCERT, p. 12). Alternatively, you can find the total number of terms and then calculate the (n – k + 1)th term.
Example: Find the 8th term from the end of 5, 8, 11, …, 35
Method: Reverse the AP and use the nth term formula.
Step 1: Reverse the AP. The new first term is 35 and the new common difference is \(d = -3\).
Step 2: Apply the nth term formula for the 8th term.
\[ a_8 = 35 + (8 – 1)(-3) \]
\[ a_8 = 35 – 21 = 14 \]
Final answer: The 8th term from the end is 14.
Sum of First n Terms of an AP: Two Formulas
The chapter shows a trick Gauss used to sum 1 to 100: pair the first and last terms, which gives a constant sum. Repeating this for n terms yields the first sum formula for an AP:
\[ S_n = \frac{n}{2}[2a + (n-1)d] \]
If you know the first term \(a\) and the last term \(l\), you can use the second sum formula:
\[ S_n = \frac{n}{2}(a + l) \]
The second form is useful when the last term is given but the common difference is unknown. Here, \(S_n\) represents the sum, \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term (NCERT, p. 16).
Solved Examples Using the Sum Formula
Example 1: Find the sum of the first 18 terms of an AP
Method: Use the sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\).
Step 1: Substitute \(a = 11\), \(d = 4\), and \(n = 18\).
\[ S_{18} = \frac{18}{2}[2(11) + (18 – 1)(4)] \]
Step 2: Simplify the bracket.
\[ S_{18} = 9[22 + 17(4)] = 9[22 + 68] = 9[90] \]
Final answer: The sum of the first 18 terms is 810.
Example 2: Find the sum given a=2, l=38, n=10
Method: Use the sum formula \(S_n = \frac{n}{2}(a + l)\) since the last term is known.
Step 1: Substitute \(a = 2\), \(l = 38\), and \(n = 10\).
\[ S_{10} = \frac{10}{2}(2 + 38) \]
Step 2: Simplify the equation.
\[ S_{10} = 5 \times 40 = 200 \]
Final answer: The sum of the 10 terms is 200.
Example 3: How many terms of 15, 12, 9, … give a sum of 54?
Method: Equate the sum formula to 54, form a quadratic equation, and find n.
Step 1: Identify \(a = 15\), \(d = -3\), and \(S_n = 54\). Substitute into the sum formula.
\[ 54 = \frac{n}{2}[2(15) + (n – 1)(-3)] \]
Step 2: Simplify to form a quadratic equation.
\[ 108 = n[30 – 3n + 3] \implies 108 = n[33 – 3n] \]
\[ 108 = 33n – 3n^2 \implies 3n^2 – 33n + 108 = 0 \]
Step 3: Divide by 3 and factorise.
\[ n^2 – 11n + 36 = 0 \]
\[ (n – 4)(n – 7) = 0 \implies n = 4 \text{ or } n = 7 \]
Final answer: Both values are valid. The number of terms is either 4 or 7. Two answers are possible because the AP is decreasing: the sum of the 5th to 7th terms is zero, so adding them does not change the total sum (NCERT, p. 16).
Key Formula Reference Box
| Formula | What it finds | When to use |
|---|---|---|
| \( d = a_{k+1} – a_k \) | Common difference | Checking if a list is an AP |
| \( a_n = a + (n-1)d \) | nth term | Finding a specific term or finding n for a given value |
| \( S_n = \frac{n}{2}[2a + (n-1)d] \) | Sum of n terms | When first term and common difference are known |
| \( S_n = \frac{n}{2}(a + l) \) | Sum of n terms | When first and last terms are known, but d is unknown |
| \( a_n = S_n – S_{n-1} \) | nth term from sum | When you are given a formula for the sum in terms of n |
| \( l = a + (n-1)d \) | Last term | Finding the final term of a finite AP using total count |
| \( b = \frac{a + c}{2} \) | Arithmetic mean | Finding the middle value if a, b, c are in AP (p. 24) |
Common Mistakes Students Make in This Chapter
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Subtracting \(a_k – a_{k+1}\) for a decreasing AP, getting a positive d (p. 4) | Always subtract the kth term from the (k+1)th term: \(d = a_{k+1} – a_k\) | For 10, 7, 4, \(d = 7 – 10 = -3\) Confirm d is negative for decreasing lists |
| Writing the nth term formula as \(a + nd\) | The correct term is \(a_n = a + (n-1)d\) | Test with the 1st term: put \(n=1\) \(a_1 = a + (1-1)d = a\), which must equal the first term |
| Reporting n as a fraction and still calling it a term | n MUST be a positive integer to be a valid term position | If you get \(n = \frac{15}{3}\), it is not a term; the denominator must divide perfectly |
| Discarding one root of the quadratic in sum problems | Both positive integer roots of n can be valid if terms cancel out (p. 16) | Plug both values of n back into the sum formula to see if both yield the required sum |
| Confusing \(S_n\) with \(a_n\) | \(S_n\) is the total sum, \(a_n\) is a single term. \(a_n = S_n – S_{n-1}\) | Check the question wording; ‘sum of’ implies \(S_n\), ‘term’ implies \(a_n\) |
Exam Notes: What the Examiner Checks
Based on the exercise structure, 3-mark questions often ask you to find the common difference and the nth term given two conditions (Exercise 5.2 style). Writing the AP in its general form earns a method mark.
4-mark questions frequently involve finding n when a sum is given, which produces a quadratic equation. Word problems about salaries, instalments, or tree-planting are common. In sum problems that yield two positive integer roots for n, writing the reason “both values are valid as the terms cancel out” earns the final reasoning mark (NCERT, p. 16).
Always state the values of a and d explicitly before substituting them into a formula. This secures the first step mark even if your calculation later has an error.
Real-Life Applications of Arithmetic Progressions
APs appear in many practical situations. A ladder has rungs that decrease uniformly in length, meaning the common difference is constant and negative. If the bottom rung is 45 cm and the next is 43 cm, the lengths form a finite AP.
Simple Interest vs Compound Interest: If ₹ 1000 is invested at 8% simple interest, the interest earned each year is ₹ 80, ₹ 160, ₹ 240… This is an AP with a common difference of 80 (NCERT, p. 11). However, if the same money is invested at 8% compound interest, the amounts become 1080, 1166.4, 1259.71… This is not an AP because compound interest multiplies the amount by a fixed factor, it does not add a fixed difference.
Counting numbers divisible by a fixed number: To find how many terms lie in a range, write the list. For two-digit multiples of 7, the list is 14, 21, 28, …, 98 (NCERT, p. 11 example). Here, the first term \(a = 14\), the common difference \(d = 7\), and the last term \(a_n = 98\). You can then use the nth term formula to solve for n and find how many such numbers exist. To solve real-world problems that use APs alongside an unknown variable, reviewing Quadratic Equations notes is helpful. For spatial problems involving steps or ladder rungs, our Triangles notes may also be useful.
Quick Revision Recap of the Chapter
- AP Definition: A list of numbers where each term is found by adding a fixed number \(d\) to the preceding term.
- d Check: Always find \(d\) using \(d = a_{k+1} – a_k\). It can be positive, negative, or zero.
- nth Term: \(a_n = a + (n-1)d\) to find any term or to check if a number belongs to the AP.
- Sum Formulas: Use \(S_n = \frac{n}{2}[2a + (n-1)d]\) or \(S_n = \frac{n}{2}(a + l)\) if the last term \(l\) is known.
- Integer Rule: n must always be a positive integer. Fractional n means the number is not a term of that AP.
Frequently Asked Questions on Arithmetic Progressions
Can the common difference of an AP be zero or negative?
Yes. The common difference \(d\) can be zero, resulting in a constant AP like 5, 5, 5, 5… It can also be negative, resulting in a decreasing AP like 10, 7, 4, 1… where \(d = -3\).
Which sum formula should I use when the last term is given but the common difference is not?
Use the second sum formula: \(S_n = \frac{n}{2}(a + l)\), where \(a\) is the first term and \(l\) is the last term. You avoid the \(d\) entirely, making it perfect for cases where the common difference is unknown.
When I solve a sum problem and get two values of n, are both correct?
Yes, if both values are positive integers. In a decreasing AP where terms go from positive to negative, the sum can reach a specific value, then later reach that exact same value again because the positive and negative terms cancel each other out.
If I get a fraction value of n while checking if a number is a term, what does it mean?
It means the number is not a term of that AP. The position of a term \(n\) must be a positive whole number (1, 2, 3…). If solving gives a fraction, the given number does not appear in that sequence.
What is the difference between the nth term and the sum of the first n terms?
The nth term \(a_n\) is a single number at a specific position in the sequence. The sum of the first n terms \(S_n\) is the total value obtained by adding the first n terms together. They are related by the formula \(a_n = S_n – S_{n-1}\).
Reference: NCERT Class 10 Mathematics textbook, chapter Arithmetic Progressions. For official curriculum verification, visit the NCERT Mathematics textbook chapter list. Also see our main Class 10 notes page for more subjects.
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