If a base 1=139/3, a base 2= -19 and a base k= a base k-1 - a base k-2, then…
2023
If a base 1=139/3, a base 2= -19 and a base k= a base k-1 - a base k-2, then what will be the 62th term in the series? (k, k-1, k-2 are in base)
- A.
139/3
- B.
-19
- C.
-196/3
- D.
-19-139/3(-19-196/3)
Attempted by 22 students.
Show answer & explanation
Correct answer: B
Key idea: Use the recurrence a_k = a_{k-1} - a_{k-2} to compute terms until a pattern repeats.
Given: a1 = 139/3, a2 = -19.
Compute a3 = a2 - a1 = -19 - 139/3 = -196/3.
Compute a4 = a3 - a2 = -196/3 - (-19) = -139/3.
Compute a5 = a4 - a3 = -139/3 - (-196/3) = 19.
Compute a6 = a5 - a4 = 19 - (-139/3) = 196/3.
Compute a7 = a6 - a5 = 196/3 - 19 = 139/3, which equals a1.
Therefore the sequence of six terms repeats: (139/3, -19, -196/3, -139/3, 19, 196/3, ...).
Since 62 ≡ 2 (mod 6), the 62nd term equals the second term, which is -19.