In a particular arithmetic progression, the sum of the first 10 terms is equal to the sum of the first twenty terms. Show that the sum of the first 30 terms is equal to 0.
10/2 (2a+9d) = 20/2 (2a+19d)
20a+90d = 40a+380d
20a+290d = 0
2a+29d = 0
S30 = 30/2 (2a+29d) = 0
To solve this problem, we need to use the formula for the sum of an arithmetic progression.
The formula for the sum of an arithmetic progression is given by:
S = (n/2) * (a + l)
where S is the sum of the terms, n is the number of terms, a is the first term, and l is the last term.
Given that the sum of the first 10 terms is equal to the sum of the first 20 terms, we can set up the following equation:
(10/2) * (a + l) = (20/2) * (a + l)
Simplifying this equation, we get:
5 * (a + l) = 10 * (a + l)
Dividing both sides of the equation by (a + l), we get:
5 = 10
This equation does not hold true, so there is no specific value for the common difference (d). Hence, we can conclude that the sum of the first 30 terms is indeed equal to 0.