The problem has two parts. Part 1: Given a quadratic equation $7x^2 - 2x + 1 = 0$ with roots $a$ and $b$, we need to form a quadratic equation with integral coefficients whose roots are $\frac{1}{a}$ and $\frac{1}{b}$. Part 2: Given three consecutive numbers $p-4$, $p+2$, and $3p+1$ in a geometric progression (G.P.), we need to find the two possible values of the common ratio of the G.P.

1. Problem Description

The problem has two parts.

Part 1: Given a quadratic equation $7x^2 - 2x + 1 = 0$ with roots $a$ and $b$, we need to form a quadratic equation with integral coefficients whose roots are $\frac{1}{a}$ and $\frac{1}{b}$.

Part 2: Given three consecutive numbers $p-4$, $p+2$, and $3p+1$ in a geometric progression (G.P.), we need to find the two possible values of the common ratio of the G.P.

2. Solution Steps

Part 1:

For the quadratic equation $7x^2 - 2x + 1 = 0$, the sum of the roots is $a+b = -\frac{-2}{7} = \frac{2}{7}$ and the product of the roots is $ab = \frac{1}{7}$.

We need to find a quadratic equation whose roots are $\frac{1}{a}$ and $\frac{1}{b}$.

The sum of the new roots is $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} = \frac{\frac{2}{7}}{\frac{1}{7}} = 2$.

The product of the new roots is $\frac{1}{a} \cdot \frac{1}{b} = \frac{1}{ab} = \frac{1}{\frac{1}{7}} = 7$.

A quadratic equation with roots $\frac{1}{a}$ and $\frac{1}{b}$ can be written as

$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$

$x^2 - 2x + 7 = 0$

Part 2:

Since $p-4, p+2, 3p+1$ are in geometric progression, the ratio between consecutive terms is constant. Therefore,

$\frac{p+2}{p-4} = \frac{3p+1}{p+2}$

$(p+2)^2 = (p-4)(3p+1)$

$p^2 + 4p + 4 = 3p^2 + p - 12p - 4$

$p^2 + 4p + 4 = 3p^2 - 11p - 4$

$0 = 2p^2 - 15p - 8$

$2p^2 - 16p + p - 8 = 0$

$2p(p-8) + 1(p-8) = 0$

$(2p+1)(p-8) = 0$

Thus, $p = 8$ or $p = -\frac{1}{2}$.

Case 1: $p=8$

The numbers are $8-4=4$, $8+2=10$, $3(8)+1=25$.

The common ratio is $\frac{10}{4} = \frac{5}{2}$ and $\frac{25}{10} = \frac{5}{2}$.

So, the common ratio is $\frac{5}{2}$.

Case 2: $p=-\frac{1}{2}$

The numbers are $-\frac{1}{2} - 4 = -\frac{9}{2}$, $-\frac{1}{2} + 2 = \frac{3}{2}$, $3(-\frac{1}{2}) + 1 = -\frac{3}{2} + 1 = -\frac{1}{2}$.

The common ratio is $\frac{\frac{3}{2}}{-\frac{9}{2}} = \frac{3}{-9} = -\frac{1}{3}$ and $\frac{-\frac{1}{2}}{\frac{3}{2}} = \frac{-1}{3} = -\frac{1}{3}$.

So, the common ratio is $-\frac{1}{3}$.

3. Final Answer

Part 1: The required quadratic equation is $x^2 - 2x + 7 = 0$.

Part 2: The two possible values of the common ratio are $\frac{5}{2}$ and $-\frac{1}{3}$.