(a) \(S = \frac{n}{2} [2a + (n - 1)d]\) \(2S = n[2a + (n - 1) d]\) \(\frac{2S}{n} = 2a + (n - 1)d\) \(\frac{2S}{n} - 2a = (n - 1) d\) \(d = \frac{\frac{2S}{n} - 2a}{(n - 1)}\) = \(\frac{2S - 2an}{n(n - 1)}\) (b) (i) Given: circle ABP To prove < AOB = 2 < APB Construction : Join PO produced to Q. Proof : /OA/ = /OB/ (radii) \(\therefore x_{1} = x_{2} \) (base angles of isosceles triangle) \(\therefore < AOQ = x_{1} + x_{2}\) (exterior angle of triangle AOP) \(\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\) Similarly, < AOB = < BOQ - < AOQ. = \(2y_{2} - 2x_{2} = 2(y_{2} - x_{2}) = 2 \times < APB\) \(\therefore < AOB = 2 < APB.\) (ii) \(z = 2x \) (angle subtended at the centre) \(z = 2(126^\circ) = 252^\circ\) \(\therefore x = 360^\circ - 252^\circ = 108^\circ\) \(\therefore y = \frac{1}{2} \times 108^\circ = 54^\circ\) \(x = 108^\circ ; y = 54^\circ ; z = 252^\circ\)