If the first two numbers in a tribonacci sequence are 5 and 11 and the eighth number is 260, what is the third

1st term = 5
2nd term = 11
3rd term = x
4th term = 5 + 11 + x = x + 16
5th term = 11 + x + (x + 16) = 2x + 27
6th term = x + (x + 16) + (2x + 27) = 4x + 43
7th term = (x + 16) + (2x + 27) + (4x + 43) = 7x + 86
8th term = (2x + 27) + (4x + 43) + (7x + 86) = 13x + 156

13x + 156 = 260
13x = 104

SOLUTION: x = 8

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The sequence: 5, 11, 8, 24, 43, 75, 142, 260

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First term is 5
Second term is 11
Let third term be a
Then fourth term is 16 + a
Fifth term is 27 + 2a
Sixth term is 43 + 4a
Seventh term is 86 + 7a
Eighth term is 156 + 13a = 260 whence third term a = (260 -- 156)/13 = 8 ANSWER
Verify from 5, 11, 8, 24, 43, 75, 142, 260 as Bonacci number sequence is such that from the fourth term, each term is the sum of the preceding three terms.
T1
T2
T3
T1 + T2 + T3
T1 + 2T2 + 2T3
2T1 + 3T2 + 4T3
4T1 + 6T2 + 7T3
7T1 + 11T2 + 13T3 = 260 OR 7(5) + 11(11) + 13T3 = 260 OR 13T3 = 104 OR T3 = 8 ANSWER