1. If \((b + c - a)x = (c + a - b)y = (a + b - c)z = 2\), then show that \[ \left(\frac{1}{x} + \frac{1}{y}\right)\left(\frac{1}{y} + \frac{1}{z}\right)\left(\frac{1}{z} + \frac{1}{x}\right) = abc \]

2. If \[ \frac{a}{y + z} = \frac{b}{z + x} = \frac{c}{x + y} \] then prove that \[ \frac{a(b - c)}{y^2 - z^2} = \frac{b(c - a)}{z^2 - x^2} = \frac{c(a - b)}{x^2 - y^2} \]

3. If \[ \cfrac{a}{y + z} = \cfrac{b}{z + x} = \cfrac{c}{x + y}, \] then prove that \[ \cfrac{a(b - c)}{y^2 - z^2} = \cfrac{b(c - a)}{z^2 - x^2} = \cfrac{c(a - b)}{x^2 - y^2}. \]

4. If \(x = \sqrt{3} + \sqrt{2}\) and \(y = \frac{1}{x}\), then find the value of: \[ (x + \frac{1}{x})^2 + \left( \frac{1}{y} - y \right)^2 \]

5. If \[ \cfrac{x}{b + c - a} = \cfrac{y}{c + a - b} = \cfrac{z}{a + b - c} \] then prove that \[ (b - c)x + (c - a)y + (a - b)z = 0 \]

6. The modal class of the above frequency distribution is 15–20. So, the mode is calculated as: \[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \] Where: - \(l = 15\) (lower boundary of modal class) - \(f_1 = 28\) (frequency of modal class) - \(f_0 = 18\) (frequency of class before modal class) - \(f_2 = 17\) (frequency of class after modal class) - \(h = 5\) (class width) Substituting the values: \[ = 15 + \left(\frac{28 - 18}{2 \times 28 - 18 - 17}\right) \times 5 = 15 + \frac{10}{21} \times 5 = 15 + \frac{50}{21} = 15 + 2.38 = 17.38 \quad \text{(approx)} \] ✅ Therefore, the mode is approximately **17.38**.

7. If \(a : b = b : c\), then prove that \[ a^2 b^2 c^2 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) = a^3 + b^3 + c^3 \]

8. If \(a:b = b:c\), then prove that \[ a^2b^2c^2\left(\cfrac{1}{a^3}+\cfrac{1}{b^3}+\cfrac{1}{c^3}\right) = a^3 + b^3 + c^3 \]

9. If \[ (a+b+c)x = (b+c-a)y = (c+a-b)z = (a+b-c)w \] prove that \[ \cfrac{1}{y}+\cfrac{1}{z}+\cfrac{1}{w}=\cfrac{1}{x} \]

10. If \(a : b = b : c\), then prove that \[ \left(\cfrac{a + b}{b + c}\right)^2 = \cfrac{a^2 + b^2}{b^2 + c^2} \]

11. If \(a : b = b : c\), then prove that \[ a^2b^2c^2\left(\cfrac{1}{a^3} + \cfrac{1}{b^3} + \cfrac{1}{c^3}\right) = a^3 + b^3 + c^3 \]

12. If \(a, b, c\) are in continued geometric progression, then prove that \[ a^2 b^2 c^2 \left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) = a^3 + b^3 + c^3 \]

13. If α and β are the roots of the equation \(ax^2 + bx + c = 0\), then what is the value of \[ \left(1 + \frac{α}{β}\right)\left(1 + \frac{β}{α}\right)? \]

14. If \(a^2 = by + cz\), \(b^2 = cz + ax\), and \(c^2 = ax + by\), then prove that \[ \frac{x}{a + x} + \frac{y}{b + y} + \frac{z}{c + z} = 1 \]

15. In triangle \(\triangle ABC\), prove that: \[ \sin\left(\frac{A + B}{2}\right) + \cos\left(\frac{B + C}{2}\right) = \cos\left(\frac{C}{2}\right) + \sin\left(\frac{A}{2}\right) \]

16. If \(a, b, c\) are in continued geometric progression, prove that \[ a^2b^2c^2\left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}\right) = a^3 + b^3 + c^3 \]

17. If \(a:b = b:c\), prove that: \[ \left(\cfrac{a+b}{b+c}\right)^2 = \cfrac{a^2+b^2}{b^2+c^2} \]

18. If the current population of a village is **p**, and the annual growth rate is **2r%**, then after **n years**, the population will be: \[ \text{Population after } n \text{ years} = p \left(1 + \frac{2r}{100} \right)^n \]

(a) \(p\left(1+\cfrac{r}{100}\right)^n\) (b) \(p\left(1+\cfrac{r}{50}\right)^n\) (c) \(p\left(1+\cfrac{r}{100}\right)^{2n}\) (d) \(p\left(1-\cfrac{r}{100}\right)^n\)

19. If a cone has volume \(V\) cubic units, total surface area \(S\) square units, height \(h\) units, and base radius \(r\) units, then show that: \[ S = 2V\left(\frac{1}{h} + \frac{1}{r}\right) \]

20. If \[ \frac{x^2 - yz}{a} = \frac{y^2 - zx}{b} = \frac{z^2 - xy}{c} \] then prove that \[ (a + b + c)(x + y + z) = ax + by + cz \]

21. If \[ \frac{3 - 5x}{x} + \frac{3 - 5y}{y} + \frac{3 - 5z}{z} = 0 \] then what is the value of \[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z}? \]

22. If \(a, b, c, d\) are in continued geometric progression, then prove that \[ (b - c)^2 + (c - a)^2 + (b - d)^2 = (a - d)^2 \]

23. If \(a, b, c,\) and \(d\) are in continued geometric proportion, then prove that \[ (b - c)^2 + (c - a)^2 + (b - d)^2 = (a - d)^2 \]

24. If \[ \cfrac{x^2 - yz}{a} = \cfrac{y^2 - zx}{b} = \cfrac{z^2 - xy}{c}, \] then prove that \[ (a + b + c)(x + y + z) = ax + by + cz. \]

25. If \[ \cfrac{x^2 - yz}{a} = \cfrac{y^2 - zx}{b} = \cfrac{z^2 - xy}{c}, \] then prove that \[ (a + b + c)(x + y + z) = ax + by + cz. \]

26. For the following data of handicraft scores awarded to 45 girl students in our village, calculate the **median** of the frequency distribution:

27. If \[ x^2 : (by + cz) = y^2 : (cz + ax) = z^2 : (ax + by) = 1 \] then show that \[ \frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z} = 1 \]

28. If \(\frac{a}{b} = \frac{c}{d} = \frac{e}{f}\), then prove that each ratio equals \[ \left(\frac{la^n + mc^n + pe^n}{lb^n + md^n + pf^n}\right)^{\frac{1}{n}} \]

29. Translate of your statement in English: If the roots of the equation \((b - c)x^2 + (c - a)x + (a - b) = 0\) are equal, then prove that: \(a + c = 2b\).

30. If \(bcx = cay = abz\), then prove that \[ \frac{ax + by}{a^2 + b^2} = \frac{by + cz}{b^2 + c^2} = \frac{cz + ax}{c^2 + a^2} \]