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.

To find the values of a, b, and c, we first need to identify the wrong number in each series and determine the correct number that should replace it.

Series I: 5, 6, 8, 14, 38, 168, 878, 5918
Let us find the differences between consecutive terms:
6 - 5 = 1 = 1!
8 - 6 = 2 = 2!
14 - 8 = 6 = 3!
38 - 14 = 24 = 4!
If the pattern continues, the next difference should be 5! = 120.
So, the term after 38 should be:
38 + 120 = 158 (instead of 168).
Let's check the next terms with this pattern:
158 + 6! = 158 + 720 = 878 (which matches)
878 + 7! = 878 + 5040 = 5918 (which matches)
Therefore, the wrong number is 168, and the correct number replacing it is:
a = 158

Series II: 32, 544, 593, 809, 832, 898, 907
Let us find the differences between consecutive terms:
544 - 32 = 512 = 8³
593 - 544 = 49 = 7²
809 - 593 = 216 = 6³
Following this pattern, the next differences should be 5², 4³, and 3²:
809 + 5² = 809 + 25 = 834 (instead of 832).
Let's check the subsequent terms using 834:
834 + 4³ = 834 + 64 = 898 (which matches)
898 + 3² = 898 + 9 = 907 (which matches)
Therefore, the wrong number is 832, and the correct number replacing it is:
b = 834

Series III: 18, 38, 124, 500, 2504, 15028, 105200
Let's look at the relationship between consecutive terms:
124 * 4 + 4 = 500
500 * 5 + 4 = 2504
2504 * 6 + 4 = 15028
15028 * 7 + 4 = 105200
The general pattern is:
$T_{n+1}=T_n\times (n+1)+4$
Applying this pattern from the beginning:
18 * 2 + 4 = 40 (instead of 38).
Using 40 to check the next term:
40 * 3 + 4 = 124 (which matches)
Therefore, the wrong number is 38, and the correct number replacing it is:
c = 40

Finding y:
We are given that y is two times the highest root of the quadratic equation:
$n^2-cn+(2.5a-11)=0$
Substituting a = 158 and c = 40 into the equation:
$n^2-40n+(2.5\times 158-11)=0$
$n^2-40n+(395-11)=0$
$n^2-40n+384=0$
Solving for n:
$n^2-24n-16n+384=0$
$n(n-24)-16(n-24)=0$
$(n-24)(n-16)=0$
The roots are n = 24 and n = 16. The highest root is 24.
Since y is two times the highest root:
y = 2 * 24 = 48

Verifying the Options:
1) Option (a): (y - c) = 2³
y - c = 48 - 40 = 8. Since 2³ = 8, this statement is true.

2) Option (b): Value of b/y is more than 16
b / y = 834 / 48 = 17.375, which is indeed more than 16. This statement is true.

3) Option (c): Value of a/y is an integer
a / y = 158 / 48 = 3.29, which is not an integer. This statement is false.

Thus, both statements (a) and (b) are true.