Given below are two number series. Series I is a missing series while series II is a wrong number series which follows pattern of series I only.

1. 11, P, 181, 350, 639, 1000
2. 242, 251, 255, 280, 329, 450, 619

If 2 is added in successive terms of a series starting with the nearest prime of P, then find the 5th term of this new series
1. 67
2. 65
3. 69
4. 63

The correct answer is either 1 or 3.

Let us analyze the two number series step-by-step.

Step 1: Analyze Series I to find the pattern and the value of P
Series I is given as:
11, P, 181, 350, 639, 1000
Let us find the differences between successive terms towards the end of the series:
Difference between 1000 and 639:
$1000-639=361$
Note that $361={19}^2$.
Difference between 639 and 350:
$639-350=289$
Note that $289={17}^2$.
Difference between 350 and 181:
$350-181=169$
Note that $169={13}^2$.
The differences are the squares of prime numbers in decreasing order: ${19}^2$, ${17}^2$, ${13}^2$.
Continuing this prime pattern backwards, the primes smaller than 13 are 11, 7, 5, etc.
Thus, the preceding differences should be ${11}^2=121$ and $7^2=49$.
Let us check if this holds for the terms 11, P, and 181:
If the difference between 181 and P is ${11}^2=121$:
$P=181-121=60$.
Now let us check the difference between P and 11:
$60-11=49$, which is indeed $7^2$.
Thus, the pattern is consistent, and $P=60$.

Step 2: Determine the nearest prime of P
The value of P is 60.
The prime numbers closest to 60 are 59 (which is $60-1$) and 61 (which is $60+1$).
Both 59 and 61 are at an equal distance of 1 unit from 60. Therefore, the nearest prime of P can be either 59 or 61.

Step 3: Find the 5th term of the new series
The question states that we start with the nearest prime of P and add 2 to successive terms.
This means the difference between successive terms is 2 (an arithmetic progression with common difference $d=2$).
The formula for the n-th term of an arithmetic progression is:
$T_n=a+(n-1)d$
Here, we want to find the 5th term ($n=5$) and the common difference is $d=2$.
$T_5=a+(5-1)\times 2=a+8$

Case 1: If the starting prime is 59
First term $a=59$.
The series is: 59, 61, 63, 65, 67...
The 5th term is:
$T_5=59+8=67$ (This corresponds to statement/option 1).

Case 2: If the starting prime is 61
First term $a=61$.
The series is: 61, 63, 65, 67, 69...
The 5th term is:
$T_5=61+8=69$ (This corresponds to statement/option 3).

Therefore, the 5th term of the new series is either 1 or 3 (either 67 or 69).