Tenth Grade Mathematics: Sum of First n Terms of an AP
Calculating the cumulative total of a sequence of numbers is a cornerstone problem in mathematics. A famous mathematical legend recounts how the great German mathematician Carl Friedrich Gauss, while still a young schoolboy, astonished his teacher by instantly computing the sum of the first 100 positive integers. Gauss recognized that pairing numbers from opposite ends of an arithmetic sequence yields identical sums. This pairing principle provides the foundational derivation for the universal formula giving the sum of the first n terms of any arithmetic progression.
Derivation of the Sum Formula
Let an Arithmetic Progression have first term a and common difference d. The sum of the first n terms, denoted by Sn, is:
Sn = a + (a + d) + (a + 2*d) + ... + [a + (n - 2)*d] + [a + (n - 1)*d]
Now, write the exact same sum in reverse order, starting from the last term l = a + (n - 1)*d:
Sn = [a + (n - 1)*d] + [a + (n - 2)*d] + ... + (a + d) + a
Add these two equations vertically term by term:
Sn + Sn = [a + a + (n - 1)*d] + [(a + d) + a + (n - 2)*d] + ... + [a + (n - 1)*d + a]
2*Sn = [2*a + (n - 1)*d] + [2*a + (n - 1)*d] + ... + [2*a + (n - 1)*d]
Notice that every single column sums to the identical value: 2a + (n - 1)d! Since there are exactly n such pairs, the right side consists of n identical terms:
2*Sn = n * [2*a + (n - 1)*d]
Dividing both sides by 2 yields the celebrated primary sum formula:
Theorem (Primary Sum Formula):
n
Sn = --- * [2*a + (n - 1)*d]
2
Alternative Form Involving the Last Term
We can partition 2*a into a + a:
Sn = (n / 2) * [a + {a + (n - 1)*d}]
Recognizing that a + (n - 1)*d is the last term l (or the n-th term an):
Theorem (Secondary Sum Formula):
n
Sn = --- * [a + l]
2
This formulation states that the sum of an AP is simply the number of terms n multiplied by the average of the first and last terms: (a + l)/2.
Visualizing Gauss's Pairing for n = 100 (Sum of 1 to 100):
1 + 2 + 3 + ... + 98 + 99 + 100
100 + 99 + 98 + ... + 3 + 2 + 1
-------------------------------------------------
101 + 101 + 101 + ... + 101 + 101 + 101 (100 copies of 101)
2*S = 100 * 101 ===> S = (100 * 101) / 2 = 5050
Sum of the First n Natural Numbers
A vital corollary is the formula for the sum of the first n positive integers: 1 + 2 + 3 + ... + n. Here, first term a = 1, common difference d = 1, and last term l = n.
Sum of First n Natural Numbers:
n * (n + 1)
Sn = ---------------
2
Extracting the n-th Term from the Sum Formula
If a formula for the sum of the first n terms Sn is given, the individual n-th term an can be obtained by subtracting the sum of the first (n - 1) terms:
Theorem:
an = Sn - S_(n-1), for n >= 2
With initial term: a1 = S1
Example: If Sn = 3*n^2 + 5*n, find the AP.
Step 1: Find a1 = S1 = 3*(1)^2 + 5*(1) = 8.
Step 2: Find S2 = 3*(2)^2 + 5*(2) = 12 + 10 = 22.
Step 3: Find a2 = S2 - S1 = 22 - 8 = 14.
Step 4: Common difference d = a2 - a1 = 14 - 8 = 6.
Conclusion: The AP is 8, 14, 20, 26, ...
Chapter Exercises
Exercise 1: Find the sum of the first 22 terms of the AP: 8, 3, -2, ...
Exercise 2: How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78? Explain why we obtain two answers.
Exercise 3: If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of the first n terms.
Solutions and Explanations
Solution 1: For the AP 8, 3, -2, ..., first term a = 8, common difference d = 3 - 8 = -5, and number of terms n = 22. Using the sum formula: S22 = (n/2) * [2a + (n - 1)d] = (22/2) * [2(8) + (22 - 1)(-5)] = 11 * [16 + 21*(-5)] = 11 * [16 - 105] = 11 * (-89) = -979. Thus, the sum of the first 22 terms is -979.
Solution 2: For the AP 24, 21, 18, ..., first term a = 24, common difference d = 21 - 24 = -3. We are given Sn = 78. Substitute into formula: (n/2) * [2(24) + (n - 1)(-3)] = 78 ==> (n/2) * [48 - 3n + 3] = 78 ==> n * (51 - 3n) = 156 ==> 51n - 3n^2 = 156. Divide by 3: 17n - n^2 = 52 ==> n^2 - 17n + 52 = 0. Factor this quadratic: find factors of 52 that add to -17; these are -13 and -4. (n - 4)*(n - 13) = 0 ==> n = 4 or n = 13. Both are positive integers. The physical meaning of two valid answers is that the common difference is negative (-3). The first 4 terms are 24, 21, 18, 15, which sum to 78. The subsequent terms from the 5th to the 13th term are positive, zero, and negative (12, 9, 6, 3, 0, -3, -6, -9, -12); these additional 9 terms sum to exactly zero (12 - 12 + 9 - 9 + 6 - 6 + 3 - 3 + 0 = 0). Thus, their inclusion does not alter the cumulative sum of 78.
Solution 3: We are given S7 = 49 and S17 = 289. Using the sum formula for n = 7: (7/2) * [2a + 6d] = 49 ==> 7 * (a + 3d) = 49 ==> a + 3d = 7 (Equation 1). For n = 17: (17/2) * [2a + 16d] = 289 ==> 17 * (a + 8d) = 289 ==> a + 8d = 17 (Equation 2). Subtract Equation 1 from Equation 2: (a + 8d) - (a + 3d) = 17 - 7 ==> 5d = 10 ==> d = 2. Substitute d = 2 into Equation 1: a + 3(2) = 7 ==> a + 6 = 7 ==> a = 1. Now find the sum of the first n terms: Sn = (n/2) * [2a + (n - 1)d] = (n/2) * [2(1) + (n - 1)(2)] = (n/2) * [2 + 2n - 2] = (n/2) * (2n) = n^2. Thus, the sum of the first n terms is n^2.