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.