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.
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
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