Given:

​(Left) U C & T % 7 D 5 4 @ 6 K & 2 R Y # 3 E * S (Right)

Logic: Letter – Symbol – Number.

(Left) U C & T % 7 D 5 4 @ 6 K & 2 R Y # 3 E * S (Right)

So, such Three symbols are there each of which is immediately preceded by a letter and also immediately followed by a number.

Thus, the correct option is: (c)

Given:

(Left) 2 R @ 4 T ! 6 # 8 U 7 % 5 & * Y H + 3 PA (Right)

Logic: Number | Symbol | Number

So, Two symbols are immediately preceded by a number and also immediately followed by a number.

Thus, the correct option is: (c)

Given:

​(Left) 2 / \ < > 2 5 $ % + 6 3 / 6 5 6 9 3 @ < & (Right)

Logic: Number | Number | Symbol

(Left) 2 / \ < > 2 5 $ % + 6 3 / 6 5 6 9 3 @ < & (Right)

So, such Three numbers are there each of which is immediately preceded by another number and also immediately followed by a symbol.

Thus, the correct option is: (d)

Given:

​(Left) 4 7 + $ £ 3 & 8 5 @ 1 Ω # * % 6 2 9 (Right)

Logic: Number | Number | Number

(Left) 4 7 + $ £ 3 & 8 5 @ 1 Ω # * % 6 2 9 (Right

So, such One numbers are there each of which is immediately preceded by another number and also immediately followed by another number.

Thus, the correct option is: (c)

Given:

​(Left) + # 5 3 $ Ω 4 ? 9 1 & % 7 * @ 2 £ 6 8 (Right)

Logic: Symbol | Symbol | Even Number

(Left) + # 5 3 $ Ω 4 ? 9 1 & % 7 * @ 2 £ 6 8 (Right)​

So, such Two symbols are there, each of which is immediately preceded by a symbol and also immediately followed by an even number.

Thus, the correct option is: (d)

Given:

​(Left) 0 3 # π 2 σ Ω 9 $ # 4 8 9 % 1 & 5 @ 8 6 μ (Right)

Logic: Prime Number | Symbol | Symbol

​(Left) 0 3 # π 2 σ Ω 9 $ # 4 8 9 % 1 & 5 @ 8 6 μ (Right)

So, such Two symbols are there, which are immediately preceded by a prime number and also immediately followed by a symbol.

Thus, the correct option is: (b)

Series:

& 1 # 6 * ^ @ 3 & $ 7 Ω $ 2 # 6 @ 2 % 9

Logic:

Remove all symbols → keep only digits → find 3rd from left.

Explanation:

Digits in order:

1, 6, 3, 7, 2, 6, 2, 9

Third from left = 3

Final Answer: 3

Final Correct Option: B

Given series: R 4 % T 9 K 2 A # 7 P 1 M & 8 Z Q * 3 L

The 6th element from the left:

Counting from left:

1 → R
2 → 4
3 → %
4 → T
5 → 9
6 → K

The 6th element is K.

The third element to the right of K

After K:

1st right → 2
2nd right → A
3rd right → #

So, # is the third element to the right of the 6th element from left.

Thus, correct option is (b).

Given:

RK59#B2%E?A8L$14SV7!C6N@H13&D

5#2, %?8, $4V, …

Logic: Each term is formed by selecting characters in order while moving forward through the sequence. After forming one term, the next term starts a little ahead in the same sequence and follows the same selection pattern.

Explanation:

5#2 is taken from the sequence in forward order.

%?8 is formed from the next relevant characters ahead in the sequence.

$4V again follows by moving further forward.

After $4V, the next suitable characters in continuation are !, 6 and @ appearing consecutively in the forward flow of the sequence.

So, the next term is !6@.

Thus, the correct option is (d).

Given: (Left) # 4 ^ 8 8 < 2 / $ 7 $ \ 1 & ^ 8 4 \ ? \ 8 (Right)

Logic: Number | Number | Symbol

So, two such numbers are there, each of which is immediately preceded by another number and also immediately followed by a symbol.

Thus, correct option is (a).