The sum of 16th term of Ap is-504,why sum of 9th term is -126.find the sum of it we 29th term, the sum of it 20th term and the sum of it 5th term
To find the sum of the 29th term, we can use the formula for the sum of an arithmetic series:
Sn = n/2 * (2a + (n-1)d)
where Sn is the sum of the series, a is the first term, n is the number of terms, and d is the common difference.
Given that the sum of the 16th term is -504, we can substitute these values into the formula and solve for the common difference:
-504 = 16/2 * (2a + (16-1)d)
-504 = 8 * (2a + 15d)
-63 = 2a + 15d ----(1)
Similarly, using the sum of the 9th term (-126), we get:
-126 = 9/2 * (2a + (9-1)d)
-126 = 4.5 * (2a + 8d)
-28 = 2a + 8d ----(2)
Now, let's express d in terms of a using equations (1) and (2):
2a + 15d = -63
2a + 8d = -28
Subtracting equation (2) from equation (1), we get:
7d = -35
d = -5
Substituting d = -5 in equation (2), we can solve for a:
2a + 8*(-5) = -28
2a - 40 = -28
2a = 12
a = 6
Now that we have the values for a and d, we can find the sum of the 29th term (n = 29):
Sn = 29/2 * (2*6 + (29-1)*(-5))
Sn = 29/2 * (12 + 28*(-5))
Sn = 29/2 * (12 - 140)
Sn = 29/2 * -128
Sn = -1856
Therefore, the sum of the 29th term is -1856.
To find the sum of the 20th term, we substitute n = 20 into the formula:
Sn = 20/2 * (2*6 + (20-1)*(-5))
Sn = 20/2 * (12 + 19*(-5))
Sn = 10 * (-43)
Sn = -430
Therefore, the sum of the 20th term is -430.
To find the sum of the 5th term, we substitute n = 5 into the formula:
Sn = 5/2 * (2*6 + (5-1)*(-5))
Sn = 5/2 * (12 + 4*(-5))
Sn = 5/2 * (-8)
Sn = -20
Therefore, the sum of the 5th term is -20.