The 6th term of an arithmetic progression is 20 and the 11th term is -5. Find the sum of the first three term
Let's denote the first term of the arithmetic progression by 'a' and the common difference by 'd'.
From the given information, we can set up two equations:
6th term: a + 5d = 20
11th term: a + 10d = -5
We have a system of two equations with two variables, so we can solve it to find the values of 'a' and 'd'.
Subtracting the first equation from the second equation, we get:
(a + 10d) - (a + 5d) = -5 - 20
5d = -25
d = -5
Substituting this value into the first equation, we get:
a + 5(-5) = 20
a - 25 = 20
a = 20 + 25
a = 45
Therefore, the first term 'a' is 45 and the common difference 'd' is -5.
Now, we can find the sum of the first three terms by substituting these values back into the arithmetic progression formula:
Sum of the first three terms = (3/2) * [2a + (n-1)d] [where n is the number of terms]
= (3/2) * [2(45) + (3-1)(-5)]
= (3/2) * [90 + 2(-5)]
= (3/2) * [90 - 10]
= (3/2) * 80
= 120
Therefore, the sum of the first three terms is 120.