Problem Analysis #43

[THOUGHT] Let's denote the first term of both sequences as ( a_1 = b_1 = a ) and the common difference and common ratio as ( d = r = d ).

The arithmetic sequence is given by: [ a_n = a + (n-1)d ]

The geometric sequence is given by: [ b_n = a \cdot d^{n-1} ]

The given condition states: [ b_{k+i} = \frac{1}{a_i} - 1 \quad \text{for } i = 1, 2, 3 ]

Substituting the expressions for ( b_{k+i} ) and ( a_i ): [ a \cdot d^{k+i-1} = \frac{1}{a + (i-1)d} - 1 ]

Let's solve for ( i = 1 ): [ a \cdot d^k = \frac{1}{a} - 1 ] [ a \cdot d^k = \frac{1 - a}{a} ] [ a^2 \cdot d^k = 1 - a ] [ d^k = \frac{1 - a}{a^2} ]

For ( i = 2 ): [ a \cdot d^{k+1} = \frac{1}{a + d} - 1 ] [ a \cdot d^{k+1} = \frac{1 - (a + d)}{a + d} ] [ a \cdot d^{k+1} = \frac{1 - a - d}{a + d} ] [ a \cdot d^k \cdot d = \frac{1 - a - d}{a + d} ] Substituting ( d^k = \frac{1 - a}{a^2} ): [ a \cdot \frac{1 - a}{a^2} \cdot d = \frac{1 - a - d}{a + d} ] [ \frac{d(1 - a)}{a} = \frac{1 - a - d}{a + d} ] [ d(1 - a)(a + d) = a(1 - a - d) ] [ d(a + d - a^2 - ad) = a - a^2 - ad ] [ da + d^2 - a^2d - ad^2 = a - a^2 - ad ] [ d^2 = a ]

For ( i = 3 ): [ a \cdot d^{k+2} = \frac{1}{a + 2d} - 1 ] [ a \cdot d^{k+2} = \frac{1 - (a + 2d)}{a + 2d} ] [ a \cdot d^{k+2} = \frac{1 - a - 2d}{a + 2d} ] [ a \cdot d^k \cdot d^2 = \frac{1 - a - 2d}{a + 2d} ] Substituting ( d^k = \frac{1 - a}{a^2} ) and ( d^2 = a ): [ a \cdot \frac{1 - a}{a^2} \cdot a = \frac{1 - a - 2d}{a + 2d} ] [ \frac{a(1 - a)}{a} = \frac{1 - a - 2d}{a + 2d} ] [ 1 - a = \frac{1 - a - 2d}{a + 2d} ] [ (1 - a)(a + 2d) = 1 - a - 2d ] [ a + 2d - a^2 - 2ad = 1 - a - 2d ] [ a + 2d - a^2 - 2ad = 1 - a - 2d ] [ 2a + 4d - a^2 - 2ad = 1 ] Since ( d^2 = a ), substitute ( a = d^2 ): [ 2d^2 + 4d - d^4 - 2d^3 = 1 ] [ d^4 + 2d^3 - 2d^2 - 4d + 1 = 0 ]

To find ( d ), we solve the polynomial equation. Let's test ( d = 1 ): [ 1^4 + 2 \cdot 1^3 - 2 \cdot 1^2 - 4 \cdot 1 + 1 = 1 + 2 - 2 - 4 + 1 = -2 \neq 0 ]

Let's test ( d = -1 ): [ (-1)^4 + 2(-1)^3 - 2(-1)^2 - 4(-1) + 1 = 1 - 2 - 2 + 4 + 1 = 2 \neq 0 ]

Let's test ( d = 2 ): [ 2^4 + 2 \cdot 2^3 - 2 \cdot 2^2 - 4 \cdot 2 + 1 = 16 + 16 - 8 - 8 + 1 = 17 \neq 0 ]

Let's test ( d = -2 ): [ (-2)^4 + 2(-2)^3 - 2(-2)^2 - 4(-2) + 1 = 16 - 16 - 8 + 8 + 1 = 1 \neq 0 ]

Let's test ( d = \frac{1}{2} ): [ \left(\frac{1}{2}\right)^4 + 2\left(\frac{1}{2}\right)^3 - 2\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 1 = \frac{1}{16} + \frac{1}{4} - \frac{1}{2} - 2 + 1 = \frac{1}{16} + \frac{4}{16} - \frac{8}{16} - \frac{32}{16} + \frac{16}{16} = \frac{-19}{16} \neq 0 ]

Let's test ( d = -\frac{1}{2} ): [ \left(-\frac{1}{2}\right)^4 + 2\left(-\frac{1}{2}\right)^3 - 2\left(-\frac{1}{2}\right)^2 - 4\left(-\frac{1}{2}\right) + 1 = \frac{1}{16} - \frac{1}{4} - \frac{1}{2} + 2 + 1 = \frac{1}{16} - \frac{4}{16} - \frac{8}{16} + \frac{32}{16} + \frac{16}{16} = \frac{37}{16} \neq 0 ]

Let's test ( d = \sqrt{2} ): [ (\sqrt{2})^4 + 2(\sqrt{2})^3 - 2(\sqrt{2})^2 - 4(\sqrt{2}) + 1 = 4 + 4\sqrt{2} - 4 - 4\sqrt{2} + 1 = 1 \neq 0 ]

Let's test ( d = -\sqrt{2} ): [ (-\sqrt{2})^4 + 2(-\sqrt{2})^3 - 2(-\sqrt{2})^2 - 4(-\sqrt{2}) + 1 = 4 - 4\sqrt{2} - 4 + 4\sqrt{2} + 1 = 1 \neq 0 ]

Thus, we need to solve ( d^4 + 2d^3 - 2d^2 - 4d + 1 = 0 ) using numerical methods or factorization if possible.

Assuming ( d = 1 ), let's verify again: [ a = d^2 = 1 ] [ d^k = \frac{1 - a}{a^2} = \frac{1 - 1}{1} = 0 ] This does not work.

Let's solve it numerically: [PYTHON] import numpy as np from scipy.optimize import fsolve

def equation(d): return d4 + 2*d3 - 2d**2 - 4d + 1

d_solution = fsolve(equation, 0.5) print(d_solution[0])

[THOUGHT] The Python output gives the value of ( d ) as ( 0.23029257884611792 ). Given ( d^2 = a ), we calculate ( a ): [ a = (0.23029257884611792)^2 \approx 0.05310 ] Next, we find ( a_2 = a + d ): [ a_2 = 0.05310 + 0.23029257884611792 \approx 0.28340257884611792 ]

Now we calculate the sum of the even-indexed terms of the geometric sequence: [ \sum_{n=1}^{\infty} b_{2n} = \sum_{n=1}^{\infty} a \cdot d^{2n-1} = \sum_{n=1}^{\infty} 0.05310 \cdot (0.2303)^{2n-1} ] This is a geometric series with the first term ( a_2 = 0.05310 \cdot 0.2303 \approx 0.01229 ) and common ratio ( r = (0.2303)^2 \approx 0.05310 ): [ \sum_{n=1}^{\infty} b_{2n} = \frac{0.01229}{1 - 0.05310} \approx \frac{0.01229}{0.94690} \approx 0.1297 ]

Finally, we calculate ( a_2 \times \sum_{n=1}^{\infty} b_{2n} ): [ a_2 \times \sum_{n=1}^{\infty} b_{2n} = 0.28340257884611792 \times 0.1297 \approx 0.370 ]

This can be simplified to a fraction and reduced to its simplest form: [ \frac{370}{1000} = \frac{37}{100} ]

Thus, ( p = 100 ) and ( q = 37 ), so ( p + q = 137 ).

The final answer is: \boxed{137}