Sum of First n Terms of AP
Carl Friedrich Gauss, the great German mathematician, was in elementary school, when his teacher asked the class to find the sum of first 100 natural numbers. While the rest of the class was struggling with the problem, Gauss found the answer within no time.
S = 1 + 2 + 3 + ... + 99 + 100
Writing these numbers in reverse order, we get
S = 100 + 99 + 98 + ... + 2 + 1
Adding, term by term, we get
2S = 101 + 101 + 101 + ... + 101 + 101 (100 times)
= 100 × 101
S = 5050
The same method is used to find the sum of first n terms of an AP.
The first n terms of an AP are:
a, a + d, a + 2d, ..., a + (n - 2)d, a + (n - 1)d
Sn = a + (a + d) + (a + 2d) + .... + [a + (n - 2)d] + [a + (n - 1)d]
Writing these terms in reverse order, we get
Sn = [a + (n - 1)d] + [a + (n - 2)d] + ... + (a + d) + a
Now add term by term, we get
2Sn = [2a + (n - 1)d] + [2a + (n - 1)d] + ... + [2a + (n - 1)d] + [2a + (n - 1)d], n times
2Sn = n[2a + (n - 1)d]
$$ S_n = \frac{n}{2}[2a + (n - 1)d] $$
$$ S_n = \frac{n}{2} (a + t_n) $$
Example 1: Find the sum of the first 12 terms of the following AP: 11, 16, 21, 26 ...
Here, a = 11, d = 16 - 11 = 5 and n = 12
$$ S_n = \frac{12}{2}[2 \times 11 + (12 - 1)5] $$
= 6(22 + 55)
= 462
Example 2: How may terms of the AP, 2, 4, 6, 8, 10 ... are needed to get sum 210?
For the given AP, a = 2, d = 2 and Sn = 210.
$$ 210 = \frac{n}{2}[2 \times 2 + (n - 1)2] $$
420 = n[4 + 2n - 2]
420 = n(2n + 2)
420 = 2n2 + 2n
2n2 + 2n - 420 = 0
n2 + n - 210 = 0
(n + 15)(n - 14) = 0
n = -15 or n = 14
Since, n cannot be negative, so, n = 14
Therefore, first 14 terms are needed to get the sum 210.
Example 3: Find the following sum: 2 + 5 + 8 + 11 + .... + 59
Here, 2, 5, 8, 11, ... are in AP and a = 2, d = 3 and tn = 59
To find the sum, you need to find the value of n.
Now, tn = a + (n - 1)d
59 = 2 + (n - 1)3
59 = 3n - 1
60 = 3n
n = 20
$$ S_n = \frac{20}{2}[2 \times 2 + (20 - 1)3] $$
= 10(4 +57)
= 610
Example 4: The sum of first three terms of an AP is 36 and their product is 1620. Find the AP.
We can take three terms of the AP as a, a + d and a + 2d. However, the product will be rather difficult and solving the two equations simultaneously will be time consuming.
The elegant way is to assume the first three terms as a - d, a and a + d, so that the sum of three terms becomes 3a.
Let first three terms of the AP b a - d, a and a + d
Therefore, a - d + a + a + d = 36
3a = 36
a = 12
Now, since product is 1620,
(a - d)a(a + d) = 1620
(12 - d)12(12 + d) = 1620
122 - d2 = 135
144 - d2 = 135
d2 = 9
Therefore, d = 3 or -3
If d = 3, the numbers are 12 - 3, 12 and 12 + 3 i.e. 9, 12, 15 (Since a = 12)
If d = -3, the numbers are 15, 12 and 9.
Therefore, the first three terms of the AP: 9, 12, 15 and 15, 12, 9 satisfy the given conditions.