The fifth term of the arithmetic progression -1 and the sum of the first twenty is -200, find the third term.
Let's denote the first term of the arithmetic progression as 'a' and the common difference as 'd'. Since the fifth term is -1, we have:
a + 4d = -1 (1)
The sum of the first 20 terms can be expressed as the sum of an arithmetic series:
S20 = (20/2)(2a + (20-1)d)
Since the sum is -200, we have:
(20/2)(2a + (20-1)d) = -200
10(2a + 19d) = -200
2a + 19d = -20 (2)
Solving equations (1) and (2) simultaneously, we get:
2a + 8d = -4 (3) (multiplying equation (1) by 2)
2a + 19d = -20 (2)
Subtracting equation (3) from (2):
11d = -16
d = -16/11
Substituting this value of d into equation (1), we get:
a + 4(-16/11) = -1
Multiplying both sides by 11:
11a - 64 = -11
11a = 53
a = 53/11
The third term can be expressed as:
a + 2d = (53/11) + 2(-16/11) = [53 - (32)]/11 = 21/11
Therefore, the third term is 21/11.
The sixth and seventh term of a linear sequence are respectively 23 and 48, calculate the sum of the first twenty terms of the sequence.
To find the sum of the first twenty terms of a linear sequence, we need to know the first term ('a') and the common difference ('d').
The sixth term of the sequence can be expressed as 'a + 5d', and the seventh term can be expressed as 'a + 6d'.
Given that the sixth term is 23 and the seventh term is 48, we have:
a + 5d = 23 (1)
a + 6d = 48 (2)
Subtracting equation (1) from (2), we get:
a + 6d - (a + 5d) = 48 - 23
d = 25
Substituting the value of 'd' into equation (1), we get:
a + 5(25) = 23
a + 125 = 23
a = -102
So, the first term of the sequence is -102, and the common difference is 25.
Now, the sum of the first twenty terms of the sequence can be expressed as:
S20 = (20/2)(2a + (20-1)d)
S20 = (10)(2(-102) + (19)(25))
S20 = 10(-204 + 475)
S20 = 10(271)
S20 = 2710
Therefore, the sum of the first twenty terms of the sequence is 2710.
Three consecutive term of an AP, have sum 21 and product 315, find the numbers.
Let's denote the three consecutive terms of the arithmetic progression as (a-d), a, and (a+d), where a is the middle term and d is the common difference.
Given that the sum of the three terms is 21, we have:
(a-d) + a + (a+d) = 21
3a = 21
a = 7
Now, given that the product of the three terms is 315, we have:
(a-d)(a)(a+d) = 315
(a^2 - d^2)(a) = 315
Substituting the value of a, we get:
(7^2 - d^2)(7) = 315
(49 - d^2)(7) = 315
49 - d^2 = 315/7
49 - d^2 = 45
-d^2 = -4
d^2 = 4
d = ±2
Since d is the common difference, it should be positive in an increasing arithmetic progression. Hence, d = 2.
Now, we can find the three consecutive terms:
(a-d) = 7-2 = 5
(a) = 7
(a+d) = 7+2 = 9
Therefore, the three consecutive terms of the arithmetic progression are 5, 7, and 9.
Three consecutive term of an AP, have a sum 15 and product 80 find the numbers.
Let's denote the three consecutive terms of the arithmetic progression as (a-d), a, and (a+d), where a is the middle term and d is the common difference.
Given that the sum of the three terms is 15, we have:
(a-d) + a + (a+d) = 15
3a = 15
a = 5
Now, given that the product of the three terms is 80, we have:
(a-d)(a)(a+d) = 80
(a^2 - d^2)(a) = 80
Substituting the value of a, we get:
(5^2 - d^2)(5) = 80
(25 - d^2)(5) = 80
25 - d^2 = 80/5
25 - d^2 = 16
-d^2 = -9
d^2 = 9
d = ±3
Since d is the common difference, it should be positive in an increasing arithmetic progression. Hence, d = 3.
Now, we can find the three consecutive terms:
(a-d) = 5-3 = 2
(a) = 5
(a+d) = 5+3 = 8
Therefore, the three consecutive terms of the arithmetic progression are 2, 5, and 8.