Let the first term of the arithmetic progression be 'a' and the common difference be 'd'.
The sum of the 4th and 6th terms of the arithmetic progression is 42.
The 4th term = a + 3d
The 6th term = a + 5d
Therefore, we can write the following equation:
(a + 3d) + (a + 5d) = 42
2a + 8d = 42 ------(1)
The sum of the 3rd and 9th terms of the arithmetic progression is 5r.
The 3rd term = a + 2d
The 9th term = a + 8d
Therefore, we can write the following equation:
(a + 2d) + (a + 8d) = 5r
2a + 10d = 5r ------(2)
To find the value of r, we can solve equations (1) and (2) simultaneously.
Multiplying equation (2) by 4, we get:
8a + 40d = 20r ------(3)
Subtract equation (1) from equation (3) to eliminate 'a':
(8a + 40d) - (2a + 8d) = 20r - 42
6a + 32d = 20r - 42 ------(4)
Now we have two equations:
2a + 8d = 42 ------(1)
6a + 32d = 20r - 42 ------(4)
We can solve equations (1) and (4) to find the values of 'a' and 'd'.
Multiplying equation (1) by 3, we get:
6a + 24d = 126 ------(5)
Subtracting equation (5) from equation (4) to eliminate 'd':
(6a + 32d) - (6a + 24d) = (20r - 42) - 126
8d = 20r - 168
d = (20r - 168)/8
d = 5r - 21 ------(6)
Substituting the value of 'd' from equation (6) into equation (1):
2a + 8(5r - 21) = 42
2a + 40r - 168 = 42
2a + 40r = 210
a + 20r = 105
a = 105 - 20r ------(7)
Now we have equations (6) and (7) to solve for 'a' and 'd' in terms of 'r'.
To find the 20th term, we can use the formula for the nth term of an arithmetic progression: the 20th term = a + 19d.
Substituting the value of 'a' from equation (7) and the value of 'd' from equation (6) into the formula:
20th term = (105 - 20r) + 19(5r - 21)
= 105 - 20r + 95r - 399
= 75r - 294
Therefore, the first term of the arithmetic progression is 105 - 20r, the common difference is 5r - 21, and the 20th term is 75r - 294.
The sum of the 4 and 6 terms of an A. P is 42 .the sum of the 3 and 9 terms of the terms of the progression is 5r. Find the first term , the common different and the twentieth term of je progression.
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