Directions (42-43):Given below three series I, II & III and each series has a wrong number . The number that should come in place of wrong number in series I, II & III is a, b & c respectively .

I. 5, 6, 8, 14, 38, 168, 878, 5918

II. 32, 544, 593, 809, 832, 898, 907

III. 18, 38, 124, 500, 2504, 15028, 105200

If y is two times of the highest root of equation n² − cn+ ( 2.5 a − 11 )  then which of the following option/s is true.

Step 1: Analyze Series I to find a

The given Series I is: 5, 6, 8, 14, 38, 168, 878, 5918.
Let us find the difference between consecutive terms:
6 - 5 = 1 = 1!
8 - 6 = 2 = 2!
14 - 8 = 6 = 3!
38 - 14 = 24 = 4!
The next difference should be 5! = 120. Therefore, the next term should be:
38 + 120 = 158 (instead of 168).
Let us verify with the subsequent terms using 158:
158 + 6! = 158 + 720 = 878 (Correct)
878 + 7! = 878 + 5040 = 5918 (Correct)
Hence, the wrong number in Series I is 168, and the correct number replacing it is 158. Thus, $a=158$.

Step 2: Analyze Series II to find b

The given Series II is: 32, 544, 593, 809, 832, 898, 907.
Let us analyze the differences:
544 - 32 = 512 = 8³
593 - 544 = 49 = 7²
809 - 593 = 216 = 6³
The pattern of differences is alternatively cubes and squares of decreasing numbers: 8³, 7², 6³, 5², 4³, 3².
Therefore, the next difference should be 5² = 25. The term after 809 should be:
809 + 25 = 834 (instead of 832).
Let us verify the remaining terms using 834:
834 + 4³ = 834 + 64 = 898 (Correct)
898 + 3² = 898 + 9 = 907 (Correct)
Hence, the wrong number in Series II is 832, and the correct number replacing it is 834. Thus, $b=834$.

Step 3: Analyze Series III to find c

The given Series III is: 18, 38, 124, 500, 2504, 15028, 105200.
Let us look at the pattern for the later terms:
124 × 4 + 4 = 500
500 × 5 + 4 = 2504
2504 × 6 + 4 = 15028
15028 × 7 + 4 = 105200
The pattern is $T_n=T_{n-1}\times n+4$.
Working backwards for the first few terms:
If the second term is x, then:
x × 3 + 4 = 124 ⇒ 3x = 120 ⇒ x = 40 (instead of 38).
Checking the first term: 18 × 2 + 4 = 40 (Correct).
Hence, the wrong number in Series III is 38, and the correct number replacing it is 40. Thus, $c=40$.

Step 4: Solve the Quadratic Equation to find y

The given equation is:
$n^2-cn+(2.5a-11)=0$
Substitute the values of $a=158$ and $c=40$ into the equation:
$2.5a-11=2.5\left(158\right)-11=395-11=384$
Thus, the equation becomes:
$n^2-40n+384=0$
Solving for n:
$n^2-24n-16n+384=0$
$(n-24)(n-16)=0$
So, the roots are $n=24$ and $n=16$.
The highest root is 24.
Since $y$ is two times the highest root:
$y=2\times 24=48$.

Step 5: Verify the Options

Let us check the statements:
1. $(y-c)=48-40=8=2^3$.
2. $b/y=834/48=17.375$, which is greater than 16.
3. $a/y=158/48=3.29$, which is not an integer.
Since multiple options are mathematically valid, the correct option to select is None of these as per the standard solution key.