SOLUTION: The sum of the first 4 terms of an arithmetic sequence is 24. The 7th term exceeds the 4th term by 24. Determine the sum of the first 12 terms

Question 839227: The sum of the first 4 terms of an arithmetic sequence is 24. The 7th term exceeds the 4th term by 24. Determine the sum of the first 12 terms
Answer by alicealc(293) (Show Source): You can put this solution on YOUR website!
The formula for the nth term in an arithmetic sequence is:
Un = U1 + (n - 1)d
in which Un is the nth term, U1 is the 1st term, and d is the difference.
So, for the 2nd term: U2 = U1 + (2 - 1)d = U1 + d
for the 3rd term: U3 = U1 + (3 - 1)d = U1 + 2d
for the 4th term: U4 = U1 + (4 - 1)d = U1 + 3d
and so on.

The sum of the first 4 terms of an arithmetic sequence is 24:
U1 + U2 + U3 + U4 = 24
U1 + (U1 + d) + (U1 + 2d) + (U1 + 3d) = 24
4U1 + 6d = 24
divide all by 2:
2U1 + 3d = 12

The 7th term exceeds the 4th term by 24:
7th term: U7 = U1 + (7 - 1)d = U1 + 6d
U7 = U4 + 24
U1 + 6d = U1 + 3d + 24
U1 + 6d - U1 - 3d = 24
6d = 3d + 24
6d - 3d = 24
3d = 24
d = 24/3 = 8

2U1 + 3d = 12
2U1 + 3(8) = 12
2U1 + 24 = 12
2U1 = -12
U1 = 12/2 = -6

The sum of the first n term can be found by using this formula:
Sn = n/2 * {2U1 + (n - 1)d}

so, the sum of the first 12 term is:
S12 = 12/2 * {2(-6) + (12 - 1)8}
S12 = 6 * (-12 + 11*8)
S12 = 6 * (-12 + 88)
S12 = 6 * 76
S12 = 456
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