The sum of the first twenty one terms of a linear sequence is 28 and the sum of the first twenty eight terms is 21. Find which term of the sequence is 0 and also the sum of the terms preceeding it
Let the first term of the linear sequence be a and the common difference be d.
The sum of the first n terms of a linear sequence can be calculated using the formula: Sn = n/2 * (2a + (n-1)d)
Given that the sum of the first twenty one terms is 28, we have:
28 = 21/2 * (2a + 20d)
28 = 21a + 210d ----(1)
Similarly, the sum of the first twenty eight terms is 21:
21 = 28/2 * (2a + 27d)
21 = 14a + 378d ----(2)
Multiplying equation (1) by 14 and equation (2) by 21, we get:
392 = 294a + 2940d ----(3)
441 = 294a + 7938d ----(4)
Subtracting equation (3) from equation (4), we have:
49 = 5598d
Therefore, d = 49 / 5598 = 1/114
Substitute d back into equation (1), we have:
28 = 21a + 210(1/114)
28 = 21a + 35/57
1470 = 1197a + 35
Solving for a:
1470 - 35 = 1197a
1435 = 1197a
a = 1435 / 1197 = 1435 / 1197 = 35 / 29
The linear sequence is:
a, a+d, a+2d, a+3d, ...
Substitute a and d back into the sequence:
35/29, 35/29 + 1/114, 35/29 + 2/114, 35/29 + 3/114, ...
To find which term of the sequence is 0, we solve for n in the equation a + (n-1)d = 0:
35/29 + (n-1)*(1/114) = 0
35/29 + (n-1)*(1/114) = 0
n = 25
The term that is equal to 0 is the 25th term in the sequence.
To find the sum of the terms preceding it (24 terms), we use the formula for the sum of n terms:
S24 = 24/2 * (2(35/29) + (24-1)*(1/114))
S24 = 12 * (70/29 + 23/114)
S24 = 12 * (1618/494)
S24 = 12 * (809/247)
S24 = 2427/247
Therefore, the sum of the terms preceding the term that is 0 is 2427/247.