Solution:
Work column-wise with a constant step of −8 in alphabet positions (A=1,…,Z=26):
Col1: Z(26) → R(18) → ? ⇒ 26−8=18−8=10 ⇒ J
Col2: ? → O(15) → G(7) ⇒ top must be 23 ⇒ W
Col3: S(19) → ? → C(3) ⇒ middle must be 11 ⇒ K
Thus the three letters (top-middle, middle-right, bottom-left) are W, K, J → “WKJ”.
Left letters $A,B,D,G,K$ with steps $+1,+2,+3,+4$ → next $+5$: $K$ stays as left? (already K from prev step), actually pattern gives next pair $K\to P$ and right $B,D,G,K\to P$; number $1,2,3,4\to5$.
So $K5P$.
$\boxed{K_5P}$
First letters $A,B,D,G$ with jumps $+1,+2,+3$ → next $+4$: $K$.
Second letters $B,D,F,H$ ($+2$ each) → $J$.
Third letters $Z,Y,X,W$ ($-1$ each) → $V$.
$\boxed{\text{KJV}}$
Solution:
Look at odd and even positions separately.
Odd positions (1,3,5,7,9,11,13): b, a, ?, b, ?, b, ? → pattern = "bab" repeating ⇒ 5=b, 9=a, 13=b.
Even positions (2,4,6,8,10,12,14): a, b, a, a, b, a, ? → pattern = "aba" repeating ⇒ 14=a.
Missing terms (in order 5th, 9th, 13th, 14th) = b, a, b, a → **baba**.
Solution:
Two interleaved sequences:
- 1st,3rd,5th,7th,…: Z(26), X(24), V(22), T(20) ⇒ decreasing by 2 ⇒ next = R(18).
- 2nd,4th,6th,8th,…: L(12), J(10), H(8), F(6) ⇒ decreasing by 2 ⇒ next = D(4).
So the next two letters are **R, D**.
Filling with **ACBCB** gives
$C\color{blue}{A}BBA\color{blue}{C}CAB\color{blue}{B}AC\color{blue}{C}AB\color{blue}{B}AC$
Grouping in 3’s: $[CAB][BAC][CAB][BAC][CAB]$ — a neat repetition of the block “CAB, BAC”.
Other choices break this pattern.