1. Problem a: Given $a^x = b^y = c^z$ and $a, b, c$ are in geometric progression (G.P.), prove that $x, y, z$ are in harmonic progression (H.P.). Step 1: Since $a, b, c$ are in G.P., there exists a common ratio $r$ such that $b = ar$ and $c = ar^2$. Step 2: Given $a^x = b^y = c^z = k$ (some constant $k$). Step 3: Taking natural logarithms, $$x \ln a = y \ln b = z \ln c = \ln k$$ Step 4: Express $y$ and $z$ in terms of $x$: $$y = \frac{\ln k}{\ln b} = \frac{x \ln a}{\ln (a r)} = \frac{x \ln a}{\ln a + \ln r}$$ $$z = \frac{\ln k}{\ln c} = \frac{x \ln a}{\ln (a r^2)} = \frac{x \ln a}{\ln a + 2 \ln r}$$ Step 5: Set $p = \ln a$ and $q = \ln r$ for simplicity, then $$x = \frac{\ln k}{p}, \quad y = \frac{\ln k}{p + q}, \quad z = \frac{\ln k}{p + 2q}$$ Step 6: Since $x, y, z$ are given as above, consider their reciprocals: $$\frac{1}{x} = \frac{p}{\ln k}, \quad \frac{1}{y} = \frac{p + q}{\ln k}, \quad \frac{1}{z} = \frac{p + 2q}{\ln k}$$ Step 7: The terms $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ form an arithmetic progression (A.P.) with common difference $\frac{q}{\ln k}$ since $$2 \times \frac{1}{y} = \frac{1}{x} + \frac{1}{z}$$ Step 8: Therefore, $x, y, z$ are in harmonic progression (H.P.). 2. Problem b: Given matrix $A = \begin{bmatrix} 2x - 1 & x \\ x + 21 & 0 \end{bmatrix}$ is symmetric, find $x$. Step 1: Symmetric means $A = A^T$, so the element at position $(1, 2)$ equals that at $(2, 1)$: $$x = x + 21$$ Step 2: This implies $$x = x + 21 \implies 21 = 0$$ Step 3: This is a contradiction, so the problem likely has a typo or needs clarification. Assuming intended second element is $x + 21 = x$ for symmetry, then $$x + 21 = x \implies 21 = 0$$ No solution unless $21=0$, which is false. Step 4: Alternatively, if $x + 21$ should be $x - 21$, then: $$x = x - 21 \implies 0 = -21$$ Still no solution. Step 5: If we assume the second element in the first row is $x$ and in second row first column is also $x$, matrix is symmetric without condition. Step 6: So value of $x$ is not restricted by symmetry unless the entries differ. 3. Problem c: Find equation whose roots are cubes of roots of $x^2 - x - 6 = 0$. Step 1: Given equation: $$x^2 - x - 6 = 0$$ Roots $\alpha, \beta$ satisfy $$\alpha + \beta = 1$$ $$\alpha \beta = -6$$ Step 2: We want an equation with roots $\alpha^3$ and $\beta^3$. Step 3: Compute $S = \alpha^3 + \beta^3$ and $P = \alpha^3 \beta^3$. Step 4: Using identity: $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3 \alpha \beta (\alpha + \beta)$$ $$= 1^3 - 3 (-6)(1) = 1 + 18 = 19$$ Step 5: And $$\alpha^3 \beta^3 = (\alpha \beta)^3 = (-6)^3 = -216$$ Step 6: The quadratic equation with roots $\alpha^3, \beta^3$: $$x^2 - S x + P = 0$$ $$x^2 - 19x - 216 = 0$$ Final answer: Equation is $$x^2 - 19x - 216 = 0$$