Study the following information and answer the questions that follow.
A number series is given as 20, a, b, c, d, 65
Where a, b, c and d are missing terms.
It is also given that:
I. a – 20 = (x2 + y)
II. The value of b is greater than a and the difference of b and a is equal to the
`[(x+1)^2+y]`
III. The value of c is `[(x+2)^2 + y]` more than b and the value of d is `[(x + 3)^2 +y]` more than c.
Note: x is equal to the HCF of 2 prime numbers and the value of y is equal to the smaller root of the quadratic equation `z^2 – z – 6 = 0`.
What is the value of d?

Options

A. A. 40
B. B. 38
C. C. 42
D. D. 48

Answer

C

Explanation

x = HCF of 2 prime numbers = 1 (Two prime numbers are always co-prime to each other)
`z^2-z-6=0`
`rArr z^2 - 3z +2z -6=0`
`rArr` z(z – 3) + 2(z – 3) = 0
`rArr` (z + 2)(z – 3) = 0
So, root of `z^2-z-6=0` are 3 and -2
Thus, y=-2
Given, a-20= `(x^2+y)`
`a=20 +(1^2-2)=20-1=19`
`b-a=[(x+1)^2+y]=[(1+1)^2-2]=2`
`rArr` b = a + 2 = 19 + 2 = 21
Also given,
`c=b+[(x+2)^2+y]=21+[(1+2)^2-2]=21 + 9 – 2 = 28`
`d = c + [(x + 3)^2 + y] = 28 + [(1 + 3)^2– 2] = 28 + 16 – 2 = 42`
Hence, option C is the correct answer.