Sum of the First n Terms of an AP and Its Properties for CBSE Class 10

What happens when we need to add all the terms of an Arithmetic Progression together? For instance, how could you quickly find the sum of all positive integers from $1$ to $100$? As the famous story goes, when 10-year-old Carl Friedrich Gauss was asked this question by his schoolmaster in the late 18th century, he produced the answer ($5050$) in seconds by pairing terms from opposite ends of the sequence.

In CBSE Class 10 Mathematics, Gauss's pairing technique is generalized into the Sum of the First $n$ Terms ($S_n$) of an AP. This formula is one of the most powerful algebraic tools in the syllabus, governing everything from series summation to real-world financial and physical calculations.

What You Will Learn

• Gauss's pairing insight and the algebraic derivation of the sum formula
• The primary sum formula: $S_n = \frac{n}{2}[2a + (n - 1)d]$
• The alternative first-and-last-term formula: $S_n = \frac{n}{2}(a + l)$
• The fundamental relationship between $S_n$ and the $n$-th term: $a_n = S_n - S_{n-1}$
• Standard summation formulas: sum of first $n$ natural numbers, odd integers, and even integers
• Step-by-step solved CBSE board examination problems and common pitfalls

1. Derivation of the Sum Formula

Let an AP have first term $a$, common difference $d$, and total $n$ terms. The sum of the first $n$ terms, denoted by $S_n$, is: $S_n = a + (a + d) + (a + 2d) + \dots + [a + (n - 2)d] + [a + (n - 1)d] \quad \text{--- (1)}$

Now, write the exact same sum in reverse order (starting from the last term): $S_n = [a + (n - 1)d] + [a + (n - 2)d] + \dots + (a + d) + a \quad \text{--- (2)}$

Adding Equation (1) and Equation (2) Term by Term:

Notice that adding corresponding terms vertically always produces the exact same sum:

• 1st terms: $a + [a + (n - 1)d] = 2a + (n - 1)d$
• 2nd terms: $(a + d) + [a + (n - 2)d] = 2a + (n - 1)d$
• Last terms: $[a + (n - 1)d] + a = 2a + (n - 1)d$

Since there are $n$ such pairs, adding both equations yields: $2S_n = n \times [2a + (n - 1)d]$

Dividing both sides by $2$:

The Primary Sum Formula

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

2. Alternative Formula Using the Last Term ($l$)

We can rewrite $2a$ as $a + a$: $S_n = \frac{n}{2}[a + a + (n - 1)d]$ Since the last term is $l = a_n = a + (n - 1)d$, substituting gives:

$S_n = \frac{n}{2}(a + l)$

This second formula is faster to compute whenever the first term and the last term are both known, eliminating the need to calculate the common difference $d$ first!

3. The Fundamental Relation: Finding $a_n$ from $S_n$

If the formula for the sum of $n$ terms ($S_n$) is given as an algebraic expression in $n$:

The $n$-th term $a_n$ is the difference between the sum of the first $n$ terms and the sum of the first $(n - 1)$ terms: $a_n = S_n - S_{n-1} \quad (\text{for } n > 1)$ and for the first term: $a_1 = S_1$.

Why Does This Hold?

$S_n = (a_1 + a_2 + \dots + a_{n-1}) + a_n = S_{n-1} + a_n \implies a_n = S_n - S_{n-1}$

4. Standard Summation Results

1. Sum of the First $n$ Natural Numbers: For $1 + 2 + 3 + \dots + n$, we have $a = 1, l = n$: $\Sigma n = \frac{n(1 + n)}{2} = \frac{n(n + 1)}{2}$ Example (Gauss's Problem): $S_{100} = \frac{100(101)}{2} = 50 \times 101 = 5050$.
2. Sum of the First $n$ Positive Odd Integers: For $1 + 3 + 5 + \dots + (2n - 1)$, we have $a = 1, d = 2$: $S_n = \frac{n}{2}[2(1) + (n - 1)2] = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}(2n) = n^2$ Example: Sum of first 10 odd numbers $= 10^2 = 100$.
3. Sum of the First $n$ Positive Even Integers: For $2 + 4 + 6 + \dots + 2n$, we have $a = 2, d = 2$: $S_n = \frac{n}{2}[2(2) + (n - 1)2] = \frac{n}{2}[4 + 2n - 2] = \frac{n}{2}(2n + 2) = n(n + 1)$

5. Solved CBSE Board Examination Problems

Solved Example 1: Sum of Multiples (CBSE Board Classic)

Problem: Find the sum of the first 15 multiples of 8.

Solution:

1. The multiples of 8 form an AP: $8, 16, 24, 32, \dots$
2. Here, first term $a = 8$, common difference $d = 8$, and number of terms $n = 15$.
3. Apply the sum formula: $S_{15} = \frac{15}{2}[2(8) + (15 - 1)(8)]$ $S_{15} = \frac{15}{2}[16 + 14(8)] = \frac{15}{2}[16 + 112] = \frac{15}{2}[128]$ $S_{15} = 15 \times 64 = 960$
4. Therefore, <u>the sum of the first 15 multiples of 8 is $960$</u>.

Solved Example 2: Finding $a_n$ and $d$ from $S_n$

Problem: If the sum of the first $n$ terms of an AP is given by $S_n = 3n^2 + 5n$, find its $n$-th term and hence find the 15th term.

Solution:

1. We are given $S_n = 3n^2 + 5n$.
2. To find $S_{n-1}$, replace $n$ with $(n - 1)$: $S_{n-1} = 3(n - 1)^2 + 5(n - 1) = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2$
3. Use the formula $a_n = S_n - S_{n-1}$: $a_n = (3n^2 + 5n) - (3n^2 - n - 2)$ $a_n = 3n^2 + 5n - 3n^2 + n + 2 = 6n + 2$
4. Now calculate the 15th term: $a_{15} = 6(15) + 2 = 90 + 2 = 92$
5. Therefore, <u>$a_n = 6n + 2$ and the 15th term is $92$</u>.

Solved Example 3: Finding $n$ when Sum is Given (Two Roots Explanation)

Problem: How many terms of the AP: $24, 21, 18, \dots$ must be taken so that their sum is $78$? Explain the double answer.

Solution:

1. Here $a = 24$, and $d = 21 - 24 = -3$. We are given $S_n = 78$.
2. Substitute into the sum formula: $S_n = \frac{n}{2}[2a + (n - 1)d] = 78$ $\frac{n}{2}[2(24) + (n - 1)(-3)] = 78$ $\frac{n}{2}[48 - 3n + 3] = 78$ $n(51 - 3n) = 156 \implies 51n - 3n^2 = 156$
3. Rearrange in standard quadratic form: $3n^2 - 51n + 156 = 0$ Divide throughout by 3: $n^2 - 17n + 52 = 0$
4. Factorize (product $= 52$, sum $= -17$ $\implies -13$ and $-4$): $(n - 4)(n - 13) = 0 \implies n = 4 \quad \text{or} \quad n = 13$
5. Why Are Both Answers Valid? (Double Answer Explanation):
   • The sum of the first 4 terms is: $24 + 21 + 18 + 15 = 78$.
   • The terms from the 5th to the 13th term are: $12, 9, 6, 3, 0, -3, -6, -9, -12$.
   • Notice that the sum of these terms (from 5th to 13th) is $0$ (the positive terms exactly cancel out the negative terms).
   • Therefore, the sum of 13 terms is still $78$!
6. Conclusion: <u>Both $n = 4$ and $n = 13$ are valid solutions</u>.

6. Summary and Examination Tips

Exam Tip: In questions asking to "Explain the double answer" when $S_n$ gives two positive integer values for $n$, explicitly state that "the sum of the terms from $(n_1 + 1)$ to $n_2$ is zero because positive and negative terms cancel each other out."

Common Mistake: Forgetting to write $2a$ inside the bracket. Students frequently write $\frac{n}{2}[a + (n-1)d]$, confusing $S_n$ with $a_n$! The formula has $2a$, not $a$.