The sum of 16terms oe an A.P is -504,while the sum of its 9terms is -126.Find the sum of its 30terms

sum of 16 terms is -504 ---> 8(2a + 15d) = -504

2a + 15d = -63

sum of 9 terms is -126 --> (9/2)(2a + 8d) = -126
2a + 8d = -28

subtract:
7d = -35
d=-5
in 2a+8d=-28
2a - 40 = -28
2a = 12
a=6

sum(30) = 15(12 + 29(-5)) = -1995

S16 = 16/2 * (2a+15d) = -504

S9 = 9/2 * (2a + 8d) = -126

16a + 120d = -504
9a + 36d = -126

Multiply #1 by 9 and #2 by 16, then subtract:

144a + 1080d = -4536
144a + 576d = -2016

504d = -2520
d = -5
a = 6

S30 = 15(2a+29d) = 15(12 - 145) = -1995

-1995

To find the sum of an Arithmetic Progression (A.P.), we can use the formula for the sum of the first n terms of an A.P.:

Sn = (n/2) * (2a + (n-1)d)

where Sn represents the sum of the first n terms, a is the first term, and d is the common difference.

Given that the sum of the first 16 terms is -504, we can use this information to find the value of (2a + (n-1)d) for n = 16.

-504 = (16/2) * (2a + (16-1)d)

Simplifying the equation:

-504 = 8 * (2a + 15d)
-63 = 2a + 15d [Dividing both sides by 8]

Similarly, given that the sum of the first 9 terms is -126, we can use this information to find the value of (2a + (n-1)d) for n = 9.

-126 = (9/2) * (2a + (9-1)d)
-252 = 9 * (2a + 8d)
-28 = 2a + 8d [Dividing both sides by 9]

Now we have two equations:

-63 = 2a + 15d
-28 = 2a + 8d

Solving these equations simultaneously will give us the values of a and d. Once we have the values of a and d, we can substitute them into the formula Sn = (n/2) * (2a + (n-1)d) to find the sum of the first 30 terms.