the sum of 16 terms of an A.P.is -504,while the sum of its 9 terms is -126.find the sum of its 30 terms.
16/2 (2a + 15d) = -504
9/2 (2a+8d) = -126
Solve for a and d, then find
30/2 (2a+29d)
How did you get the a and d nauu???
To find the sum of the 30 terms of an arithmetic progression (A.P.), we need to use the formula for the sum of the first n terms of an A.P.
The formula for the sum of the first n terms of an A.P. is given by: Sn = (n/2)(2a + (n-1)d), where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
Given:
The sum of the first 16 terms is -504.
The sum of the first 9 terms is -126.
Using the formula for the sum of the first 16 terms, we have:
-504 = (16/2)(2a + (16-1)d)
-504 = 8(2a + 15d)
-63 = 2a + 15d -- Eq. (1) (since 8 and -63 are divisible by 3)
Using the formula for the sum of the first 9 terms, we have:
-126 = (9/2)(2a + (9-1)d)
-126 = 4.5(2a + 8d)
-28 = 2a + 8d -- Eq. (2)
We now have a system of equations with two unknowns (a and d).
To solve for a and d, we can use either the substitution method or the elimination method. I'll use the elimination method in this case.
Multiply Eq. (1) by 4 and Eq. (2) by 15 to eliminate the variable 'a':
-252 = 8a + 60d
-420 = 30a + 120d
Subtracting the second equation from the first equation to eliminate 'a', we get:
168 = 22a + 60d
Now, we can solve for the common difference (d):
168 = 22a + 60d -- Eq. (3)
-63 = 2a + 15d -- Eq. (1) (repeating for convenience)
Multiply Eq. (1) by 30 and Eq. (3) by 15 to eliminate 'a':
-1890 = 60a + 450d
2520 = 30a + 900d
Subtracting the second equation from the first equation to eliminate 'a', we get:
4410 = 610d
Now, we can find the value of the common difference (d):
d = 4410 / 610
d ≈ 7.21311
Substituting the value of d back into Eq. (1) to find 'a':
-63 = 2a + 15d
-63 = 2a + 15(7.21311)
-63 = 2a + 108.19665
2a ≈ -171.19665
a ≈ -85.598325
Now, we have the values of the first term (a) and the common difference (d). We can calculate the sum of the first 30 terms (Sn) using the formula:
Sn = (n/2)(2a + (n-1)d)
Substituting the values:
S30 = (30/2)(2(-85.598325) + (30-1)(7.21311))
S30 ≈ 15(-171.19665 + 200.1693)
S30 ≈ 15(28.97265)
S30 ≈ 434.08975