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\(T_1\) = 153 \(T_6 = \frac{17}{27}\) The equation for the nth term is ar\(^{n-1}\) \(T_6: 153r^5\) = \(\frac{17}{27}\) divide through by 153 r\(^5\) = \(\frac{17}{27}\) x \(\frac{1}{153}\) take 5th root of both sides r = \(\frac{1}{3}\) The sum of n terms is \(S_n = \frac{a(1 - r^n)}{1 - r }\) So, \(S_4 = \frac{153(1 - (\frac{1}{3})^4)}{1 - \frac{1}{3}}\) = \(\frac{680}{3}\)

1st term a = 11, 4th term = 32 nth term = a + (n - 1)d 4th term = 11 = (4 - 1)d = 11 + 3d = 32 3d = 21 d = 7 sn = \(\frac{n}{2}\)(2a + (n - 1)d) sn = \(\frac{9}{2}\)(2 \(\times\) 11) + (9 - 1)7 \(\frac{9}{2}\)(22 + 56) = \(\frac{9}{2}\) x 78 = 351

\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 135}\) \(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 45}\) x = \(\frac{6 \times \sin 45}{\sin 30}\) = \(6 \sqrt{2}\)cm

Sum of interior angle of hexagon = [2(6) - 4]90o = 720o sum of central angle = 360o Each central angle = \(\frac{360}{6}\) = 60o Area of Hexagon = \(\frac{1}{2}\) x 6 x 6 sin 60o \(\frac{36 \times 6\sqrt{3}}{2 \times 2}\) = \(54 \sqrt{3}\)cm2

By re-arranging the goals in ascending order 0. 0. 0. 0. 0, 1. 1. 1. 1. 1. 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5. Mean = \(\frac{36}{20}\) = 1.8 Median = \(\frac{1 + 2}{2}\) = \(\frac{3}{2}\) = 1.5 Mode = 1 = 1.8, 1.5 and 1

sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3 Where 1 + tan\(^2\) \(\theta\) = sec\(^2\) \(\theta\) 1 + tan\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3 2 tan\(^2\) \(\theta\) = 2 tan\(^2\) \(\theta\) = 1 tan\(\theta\) = \sqrt1 where \sqrt1 = 1 tan\(\theta\) = 1 And tan 45^\circ = 1 ∴ \(\theta\) = 45^\circ

The rise of water is equivalent to the volume of the sphere of radius \(\frac{1}{2}\)cm immersed x \(\frac{1}{\text{No. of sides sq. root 3}}\) Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))3 = \(\frac{1}{8}\) = \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\) = \(\frac{\pi}{24\sqrt{3}}\)

x2 - y - 1 = 0.......(i) y - 2x + 2 = 0......(ii) By re-arranging eqn. (ii) y = 2x - 2........(iii) Subst. eqn. (iii) in eqn (i) x2 - (2x - 2) - 1 = 0 x2 - 2x + 1 = 0 = (x - 1) = 0 When x - 1 = 0 x = 1 Sub. for x = 1 in eqn. (iii) y = 2 - 2 = 0 x = 1, y = 0

No expanded working has been attached for this question yet. The verified answer is Option C.

(x + y) is a factor 2x3 + 2x2y - xy + 3x - y^2

Two triangles have the same area if three sides in one triangle equal to three in the other.

(25)\(^{x - 1}\) = 64(\(\frac{5}{2}\))\(^6\) (\(\frac{5}{2}\))\(^6\) = (\(\frac{5^6}{64}\)) 25\(^{x - 1}\) = 64 x (\(\frac{5^6}{64}\)) 5\(^{2x - 2}\) = 5\(^6\) 2x - 2 = 6 = \(\frac{8}{2}\) = 4

Assuming loge 4.4 = 1.4816 and loge 7.7 = 2.0142 then value of loge \(\frac{7}{4}\) = loge7.7 - loge \(\frac{7.7}{4.4}\) = loge\(\frac{7}{4}\) = 2.0142 -1.4816 = 0.5326

The locus of all points having a distance of 1 unit from each of the two fixed points. a and b is: a perpendicular to the (ab) through the mid-points ab

No expanded working has been attached for this question yet. The verified answer is Option C.