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\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\) \(T_n = ar^n - 1\) ∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i) \(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii) = 2a = 3(1 - r) = 2a = 3 - 3r ∴ a = \(\frac{3 - 3r}{2}\) Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i) = \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\) = \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\) = 3(3r - 3r\(^2\)) = -4 = 9r - 9r\(^2\) = -4 = 9r\(^2\) - 9r - 4 = 0 = 9r\(^2\) - 12r + 3r - 4 = 0 = 3r(3r - 4) + 1(3r - 4) = 0 = (3r - 4)(3r + 1) = 0 ∴ r = \(\frac{4}{3} or - \frac{1}{3}\) For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1. ∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)

Given that, a * b = a\(^2\)b and a ^ b = 2a + b (-4 * 2) = (-4)\(^2\) x 2 = 16 x 2 = 32 (7 * -1) = 7\(^2\) x (-1) = 49 x (-1) = -49 ∴ (-4 * 2) ^ (7 * -1) = 2(32) + (-49) = 64 - 49 = 15

A binary operation '*' defined on a set A is said to be commutative only if a*b=b*a, ∀a, b∈A. If (a*b)*c=a*(b*c), then the operation is said to associative ∀ a, b∈ A. If (b ο c)*a=(b*a) ο (c*a), then the operation is said to be distributive ∀ a, b, c ∈ A.

Let the length of the longer side = \(x\) cm ∴ The length of the shorter side = (\(x\) - 6) cm If we increase each side's length by 2 cm, it becomes (\(x\) + 2) cm and (\(x\) - 4) cm respectively Area of a rectangle = L x B \(A_1 = x(x - 6) = x^2 - 6x\) \(A_2 = (x + 2)(x - 4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8\) \(A_1 + 68 = A_2\) (Given) ⇒ \(x^2 - 6x + 68 = x^2 - 2x - 8\) ⇒ \(x^2 - x^2 - 6x + 2x\) = -8 - 68 ⇒ -4\(x\) = -76 ⇒ \(x\) = \(\frac{-76}{-4}\) = 19cm ∴ The length of the shorter side = \(x\) - 6 = 19 - 6 = 13 cm

Area of the composite figure = Area of semi circle + Area of rectangle + Area of triangle Area of semi circle = \(\frac{1}{2}\pi r^2 = \frac{1}{2}\times\pi\times\frac{d^2}{4} = \frac{1}{2}\times\frac{22}{7}\times\frac{42^2}{4} = 693 m^2\) Area of rectangle = l x b = 42 x 60 =2520 m\(^2\) Area of triangle = \(\frac{1}{2}\times b \times h = \frac{1}{2}\times 36 \times 42 = 756 m^2\) ∴ Area of the composite figure = 693 + 2520 + 756 = 3969 m\(^2\)

Area of trapezium = \(\frac{1}{2}(a + b) h\) ⇒ \(\frac{1}{2} (a + b)\times 16 = 200\) ⇒ 8(a + b) = 200 ⇒ a + b = \(\frac{200}{8}\) = 25 -----(i) ⇒ a : b = 2 : 3 ⇒ \(\frac{a}{b} = \frac{2}{3}\) ⇒ 3a = 2b ⇒ a = \(\frac{2b}{3}\) -------(ii) Substitute \(\frac{2b}{3}\) for a in equation (i) ⇒ \(\frac{2b}{3}\) + b = 25 \(\frac{5b}{3}\) = 25 ⇒ b = 25 \div \(\frac{5}{3} = 25\times\frac{3}{5} = 15cm\) From equation (ii) ⇒ a = \(\frac{2 \times 15}{3} = 2\times5 = 10cm\) ∴ Lengths of each parallel sides are 10cm and 15cm

Perimeter of a triangle = sum of all sides ⇒ \(P = y + x + x = 2\) ⇒ \(y + 2x = 2\) ⇒ \(y= 2 - 2x\)-----(i) Using Pythagoras theorem \(y^2 = x^2 + x^2\) ⇒ \(y^2 = 2x^2\) ⇒ \(y = \sqrt2x^2\) ⇒ \(y = x\sqrt2\)-----(ii) Equate \(y\) ⇒ \(2 - 2x = x\sqrt2\) Square both sides ⇒ \((2 -2x) ^2 = (x\sqrt2)^2\) ⇒ \(4 - 8x + 4x^2 = 2x^2\) ⇒ \(4 - 8x + 4x^2 - 2x^2 = 0\) ⇒ \(2x^2 - 8x + 4 = 0\) ⇒ \(x = \frac{-(-8)\pm\sqrt(-8)^2 - 4\times2\times4}{2\times2}\) ⇒ \(x = \frac{8\pm\sqrt32}{4}\) ⇒ \(x = \frac{8\pm4\sqrt2}{4}\) ⇒ \(x = 2\pm\sqrt2\) ⇒ \(x = 2 + \sqrt2\) or \(2 - \sqrt2\) ∴ \(x = 2 - \sqrt2\) (for \(x\) has to be less than its perimeter) ∴ \(y = 2 - 2x = 2 - 2(2 - \sqrt2) = -2 + 2 \sqrt2\) ∴ The length of the longer side = -2 + 2\(\sqrt2\)m

Using Pythagoras theorem ⇒ \((20\sqrt2)2 = x^2 + x^2\) ⇒ 800 = \(2x^2\) ⇒ 400 = \(x^2\) ⇒ x = \(\sqrt400\) = 20 cm ∴ Area of a square = \(x^2 = 20^2 = 400 cm^2\) ∴ Perimeter of a square = 4x = 4 x 20 = 80 cm

The sides have been given to the nearest meter, so 51.5 m \le length < 52.5 37.5 m \le width < 38.5 Minimum area = 37.5 x 51.5 = 1931.25 m\(^2\) Maximum area = 38.5 x 52.5 = 2021.25 m\(^2\) ∴ The range of the area = 1931.25 m\(^2\) \le A < 2021.25 m\(^2\)

Let the length of the sides of triangle be 2x, 3x and 4x. Perimeter of triangle = 180cm ⇒ \(2x +3x + 4x = 180\) ⇒ \(9x = 180\) ⇒ \(x = \frac{180}{9}\) = 20cm Then the sides of the triangle are: \(2x = 2\times20 = 40cm; 3x = 3\times20\) = 60cm and \(4x\) = 4\(\times20 \)= 80cm Using Heron's formula Area of triangle = \(\sqrt s(s-a)(s-b)(s-c)\) Where s = \(\frac{a + b + c}{2}\) Let a = 40cm, b = 60cm, c = 80cm and s = \(\frac{40 + 60 + 80}{2} = \frac{180}{2}\) = 90cm ⇒ A = \(\sqrt90 (90 - 40) (90 - 60) (90 - 80) = \sqrt90 \times 50 \times 30 \times 10 = \sqrt1350000\) ∴ A =1162cm\(^2\) (to the nearest cm\(^2)\)

Volume of the cylinder = \(\frac{\theta}{360} \times \pi r^2h\) \theta = 360\(^o\) - 90\(^o\) = 270\(^o\) ∴ Volume of the cylinder = \(\frac{270}{360} \times \frac{22}{7} \times \frac{14^2}{1} \times \frac{32}{1} = \frac{37,255,680}{2520} = 14,784cm^3\)

Volume of the composite solid = Volume of A + Volume of B Volume of a cuboid = length x breadth x height Volume of A = 6 x 26 x 8 = 1248 cm\(^3\) Volume of B = 6 x 10 x 22 = 1320 cm\(^3\) ∴ Volume of the composite solid = 1248 + 1320 = 2568 cm\(^3\)

Volume of a cone = \(\frac{1}{3}\pi r^2 h\) r = 5 cm l = 13 cm Using Pythagoras theorem ⇒ \(13^2 = 5^2 + h^2\) ⇒ \(169 = 25 + h^2\) ⇒ \(169 - 25 = h^2\) ⇒ \(h^2 = 144\) ⇒ \(h = \sqrt144 = 12 cm\) ∴ Volume of the cone = \(\frac{1}{3} \times\pi\times 5^2 x 12 = 100\pi\) cm\(^3\)

An interior angle of a regular polygon = \(\frac{(2n-4)\times 90}{n}\) An exterior angle of a regular polygon = \(\frac{360}{n}\) \(\frac{(2n-4)\times 90}{n}\) =5 \(\times\) \(\frac{360}{n}\) (Given) = (2n-4) x 90 = 5 x 360 = 180n - 360 = 1800 = 180n = 1800 + 360 = 180n = 2160 = n = \(\frac{2160}{180}\) = 12 The polygon has 12 sides which is dodecagon

The locus of a point equidistant from two intersecting lines is pair of bisectors of the angles between the two lines.