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`.
If another series follows the same pattern as the given series and the first term of this new series is 29, then find the fifth term of this new series.

Options

A. 67
B. 56
C. 42
D. 51

Explanation

To solve the problem step by step, we need to find the values of a, b, c, and d based on the given equations and then apply the same pattern to a new series starting with 29.

### Step 1: Determine the values of x and y
- **x** is the HCF of two prime numbers. The HCF of any two prime numbers is 1.
- Therefore, **x = 1**.

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

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

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

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

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

### Step 6: New series starting with 29
Now, we have the series:
- Original series: 20, 19, 21, 28, 42, 65
- New series starts with 29, so we will replace the first term (20) with 29.

The new series will be:
- First term: 29
- Second term (a): 19 + 9 = 28
- Third term (b): 21 + 9 = 30
- Fourth term (c): 28 + 9 = 37
- Fifth term (d): 42 + 9 = 51

Thus, the fifth term of the new series is **51**.

### Final Answer
The fifth term of the new series is **51**.