determine the first three terms of the erithmetic progression of which the 7th term is 3 and the 12th term is -3?
#1: a+6d=3
#2: a+11d=-3
subtract them
-5d = 6
d = -6/5
back into a+6d=3
a + 6(-6/5) = 3
a = 3 + 36/5 = 51/5
first three terms are:
51/5 , 9 , 29/5
To determine the first three terms of an arithmetic progression, we need to find the common difference between consecutive terms.
Let's start by identifying the given information:
- The 7th term is 3.
- The 12th term is -3.
We can use these two pieces of information to find the common difference.
The formula for the nth term of an arithmetic progression is given by:
term(n) = a + (n-1)d
Now we can plug in the known values:
term(7) = a + (7-1)d = 3 --> (1)
term(12) = a + (12-1)d = -3 --> (2)
We have a system of two equations with two unknowns (a and d). We can solve this system to find the values of a and d.
Let's simplify equation (1):
a + 6d = 3 --> (3)
And simplify equation (2):
a + 11d = -3 --> (4)
Now, we can solve these two equations simultaneously. Subtract equation (3) from equation (4):
(4) - (3): (a + 11d) - (a + 6d) = -3 - 3
Simplifying: 5d = -6
Dividing both sides by 5: d = -6/5
Now we have the common difference (d). We can substitute this value back into equation (3) to find a:
a + 6(-6/5) = 3
a - 36/5 = 3
a = 3 + 36/5
a = 15/5 + 36/5
a = 51/5
So we have found the values of a and d. Now we can calculate the first three terms of the arithmetic progression.
The first term, a, is 51/5.
The common difference, d, is -6/5.
Using the formula for the nth term of an arithmetic progression, we can find:
term(1) = a + (1-1)d = a
term(2) = a + (2-1)d = a + d
term(3) = a + (3-1)d = a + 2d
Plugging in the values, we get:
term(1) = 51/5
term(2) = (51/5) + (-6/5) = 45/5 = 9
term(3) = (51/5) + 2(-6/5) = 39/5
Therefore, the first three terms of the arithmetic progression are: 51/5, 9, 39/5.