The 6th term of an arithmetic progression is 20 and the 11th term is -5. Find the sum of the first three term

Answer 1

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.