The inscribed circle in the square has diameter 8 cm, so radius = 4 cm.
When a square is inscribed in this circle, its diagonal equals the diameter (8 cm).
If the diagonal of the smaller square is 8 cm, then using diagonal = side√2, we get side = 8/√2 = 4√2 cm.
Therefore, area = (4√2)² = 32 cm².
In GP: b = ar and c = ar².
Given a + ar + ar² = 39 and a² + a²r² + a²r⁴ = 651.
From the first equation: a(1 + r + r²) = 39.
From the second: a²(1 + r² + r⁴) = 651.
Dividing these equations and solving yields r = 2, which satisfies both conditions when a = 3.
From Vieta's formulas: α + β = 4/3 and αβ = k/3.
Using α² + β² = (α+β)² - 2αβ, we get 10/9 = (4/3)² - 2(k/3) = 16/9 - 2k/3.
Solving: 10/9 = 16/9 - 2k/3 gives 2k/3 = 6/9 = 2/3, so k = 1/3.
Using Vieta's formulas: p + q = 5 and pq = 6.
We need 1/p + 1/q = (p+q)/pq = 5/6.
This approach avoids finding individual roots and uses the relationship between roots directly.
When three numbers are in arithmetic progression, they can be written as (a-d), a, (a+d) where a is the middle term.
Their sum is (a-d) + a + (a+d) = 3a = 15, giving a = 5.
Therefore, the middle term y = 5.
For a quadratic equation ax² + bx + c = 0, sum of roots = -b/a and product of roots = c/a.
Here, sum = -(k-3)/2 and product = (k-5)/2.
Given that sum = (1/2) × product, we have -(k-3)/2 = (1/2) × (k-5)/2.
Solving: -(k-3)/2 = (k-5)/4, which gives -2(k-3) = k-5, leading to -2k+6 = k-5, thus k = 11.