Determining the first three terms of an ArithmetiC progressionof which the (a.)10th term is 31 and the 15th term is 49 (b.)7th term is 3 and 12th term is -3 (c.)5th term is 8 and the 11th term is -28 (d.)9th term is7+9x and d=2x,Show all calculations
To find the first three terms of an arithmetic progression, we need to use the given information about the terms.
(a.) Given: 10th term = 31, 15th term = 49
We know that the formula for the nth term of an arithmetic progression is given by:
tn = a + (n-1)d
Where tn is the nth term, a is the first term, n is the position of the term, and d is the common difference.
Let's apply this formula to the given information:
10th term (tn) = 31
tn = a + (n-1)d
31 = a + (10-1)d
31 = a + 9d -- (1)
15th term (tn) = 49
tn = a + (n-1)d
49 = a + (15-1)d
49 = a + 14d -- (2)
Now, we have two equations (1) and (2) with two variables (a and d). We can solve these equations simultaneously to find the values of a and d.
Subtracting equation (1) from equation (2), we get:
49 - 31 = a + 14d - (a + 9d)
18 = 5d
Dividing both sides by 5, we find:
d = 18/5 = 3.6
Substituting the value of d back into equation (1), we can solve for a:
31 = a + 9(3.6)
31 = a + 32.4
a = 31 - 32.4
a = -1.4
Now that we have the values of a and d, we can determine the first three terms of the arithmetic progression:
First term (a) = -1.4
Common difference (d) = 3.6
The first term (a1) is -1.4. The second term (a2) can be found by adding the common difference to the first term:
a2 = a1 + d
a2 = -1.4 + 3.6
a2 = 2.2
Similarly, the third term (a3) can be found by adding the common difference to the second term:
a3 = a2 + d
a3 = 2.2 + 3.6
a3 = 5.8
So, the first three terms of the arithmetic progression are -1.4, 2.2, 5.8.
You can follow the same steps to solve parts (b), (c), and (d) of the question.
To determine the first three terms of each arithmetic progression, we need to find the common difference (d) first, using the information given.
(a.) For the first set:
Given:
10th term (a10) = 31
15th term (a15) = 49
We know that in an arithmetic progression, the nth term (an) can be found using the formula:
an = a1 + (n-1)d
Substituting the given terms:
a10 = a1 + (10-1)d → (1)
a15 = a1 + (15-1)d → (2)
To find the common difference (d), we can subtract equation (1) from equation (2):
a15 - a10 = a1 + (15-1)d - (a1 + (10-1)d)
49 - 31 = 14d - 9d
18 = 5d
Dividing both sides by 5:
d = 18 / 5
d = 3.6
Now, we can substitute d back into equation (1) to find the first term (a1):
31 = a1 + (10-1)(3.6)
31 = a1 + 9(3.6)
31 = a1 + 32.4
a1 = 31 - 32.4
a1 = -1.4
Therefore, the first three terms of this arithmetic progression are: -1.4, 2.2, 5.8.
(b.) For the second set:
Given:
7th term (a7) = 3
12th term (a12) = -3
Using the same formula as before, we can set up two equations:
a7 = a1 + (7-1)d → (3)
a12 = a1 + (12-1)d → (4)
Subtracting equation (3) from equation (4) to find d:
a12 - a7 = a1 + (12-1)d - (a1 + (7-1)d)
-3 - 3 = 11d - 6d
-6 = 5d
Dividing both sides by 5:
d = -6 / 5
d = -1.2
Substituting d back into equation (3) to find a1:
3 = a1 + (7-1)(-1.2)
3 = a1 + 6(-1.2)
3 = a1 - 7.2
a1 = 3 + 7.2
a1 = 10.2
Therefore, the first three terms of this arithmetic progression are: 10.2, 9, 7.8.
(c.) For the third set:
Given:
5th term (a5) = 8
11th term (a11) = -28
Using the same formula as before, we can set up two equations:
a5 = a1 + (5-1)d → (5)
a11 = a1 + (11-1)d → (6)
Subtracting equation (5) from equation (6) to find d:
a11 - a5 = a1 + (11-1)d - (a1 + (5-1)d)
-28 - 8 = 10d - 4d
-36 = 6d
Dividing both sides by 6:
d = -36 / 6
d = -6
Substituting d back into equation (5) to find a1:
8 = a1 + (5-1)(-6)
8 = a1 + 4(-6)
8 = a1 - 24
a1 = 8 + 24
a1 = 32
Therefore, the first three terms of this arithmetic progression are: 32, 26, 20.
(d.) For the fourth set:
Given:
9th term (a9) = 7 + 9x
Common difference (d) = 2x
Using the nth term formula:
a9 = a1 + (9-1)d → (7)
d = 2x → (8)
Substituting equation (8) into equation (7) to find a1:
7 + 9x = a1 + (9-1)(2x)
7 + 9x = a1 + 8(2x)
7 + 9x = a1 + 16x
To simplify, we can subtract 16x from both sides:
7 - 7x = a1
Therefore, the first three terms of this arithmetic progression are: (7 - 7x), (7 - 5x), (7 - 3x).
These are the calculations for determining the first three terms of each given arithmetic progression.
(a)
a+9d = 31
a+14d = 49
d = 18/5
a = -7/5
so, we have
-7/5, 11/5, 29/5, ...
(b) and (c) are just the same.
(d)
a+8d = a+16x = 7+9x
a = 7-7x
so,
7-7x, 7-5x, 7-3x ...