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 = (x^2 + 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`.
Find the ratio between value of c and d respectively.

Options

A. `3:2`
B. `4:3`
C. `3:4`
D. `2:3`

Explanation

To solve the problem step-by-step, we will follow the given information systematically.

### Step 1: Determine the values of x and y
1. **Finding x**:
- x is the HCF of two prime numbers. The HCF of any two prime numbers is always 1, since they have no common factors other than 1.
- Therefore, \( x = 1 \).

2. **Finding y**:
- y is the smaller root of the quadratic equation \( z^2 - z - 6 = 0 \).
- To find the roots, we can factor the equation:
\[ z^2 - 3z + 2z - 6 = 0 \implies (z - 3)(z + 2) = 0 \]
- The roots are \( z = 3 \) and \( z = -2 \). The smaller root is \( y = -2 \).

### Step 2: Calculate the value of a
- From the equation \( a - 20 = x^2 + y \):
\[ a - 20 = 1^2 + (-2) = 1 - 2 = -1 \]
\[ a = 20 - 1 = 19 \]

### Step 3: Calculate the value of b
- The difference \( b - a = (x + 1)^2 + y \):
\[ b - 19 = (1 + 1)^2 + (-2) = 2^2 - 2 = 4 - 2 = 2 \]
\[ b = 19 + 2 = 21 \]

### Step 4: Calculate the value of c
- The value of c is given by \( c = b + (x + 2)^2 + y \):
\[ c = 21 + (1 + 2)^2 + (-2) = 21 + 3^2 - 2 = 21 + 9 - 2 = 21 + 7 = 28 \]

### Step 5: Calculate the value of d
- The value of d is given by \( d = c + (x + 3)^2 + y \):
\[ d = 28 + (1 + 3)^2 + (-2) = 28 + 4^2 - 2 = 28 + 16 - 2 = 28 + 14 = 42 \]

### Step 6: Find the ratio of c to d
- The ratio of c to d is:
\[ \text{Ratio} = \frac{c}{d} = \frac{28}{42} = \frac{2}{3} \]

### Final Answer
The ratio between the value of c and d is \( \frac{2}{3} \).

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