Question

In the sequence \((\text{Left})\ S7@PB6\#!4Q\$ATKB8M\ (\text{Right})\), the following combinations are given based on the position of elements. Which of them follow the same pattern? A. AK4, B. QA!, C. BMT, D. #4P, E. 6!@.

• A. A, C, E only
• B. A, D, E only
• C. B, C, D only
• D. C, E only

Hint:

For sequence pattern questions, assign position numbers to every symbol first, then compare position differences.

Answer 1

The Correct Option is B

First write the positions of the elements in the given sequence. \[ S(1),\ 7(2),\ @(3),\ P(4),\ B(5),\ 6(6),\ \#(7),\ !(8),\ 4(9), \] \[ Q(10),\ \$(11),\ A(12),\ T(13),\ K(14),\ B(15),\ 8(16),\ M(17). \] Now check the combinations. For A. \(AK4\): \[ A=12,\quad K=14,\quad 4=9. \] The pattern is: \[ 12 \rightarrow 14 \rightarrow 9. \] So the second element is \(2\) places to the right and the third element is \(3\) places to the left of the first element. For D. \(\#4P\): \[ \#=7,\quad 4=9,\quad P=4. \] The pattern is: \[ 7 \rightarrow 9 \rightarrow 4. \] Again, the second element is \(2\) places to the right and the third element is \(3\) places to the left of the first element. For E. \(6!@\): \[ 6=6,\quad !=8,\quad @=3. \] The pattern is: \[ 6 \rightarrow 8 \rightarrow 3. \] Again, the same pattern is followed. Now check B. \(QA!\): \[ Q=10,\quad A=12,\quad !=8. \] Here the third element is \(2\) places to the left, not \(3\) places to the left. So B does not follow the same pattern. Therefore, the combinations following the same pattern are: \[ A,\ D,\ E. \]