# In an arithmetic progression, the sum of the first five...

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TJC JC1 Promotional Exams 2011_Solutions and Marker’s Comments 1 In an arithmetic progression, the sum of the first five terms is twice the sum of the next five terms. Given that the first term is 60, find the 13 th term. 
TJC JC1 Promotional Exams 2011_Solutions and Marker’s Comments 2 In the diagram, P and A are fixed points 500 m apart on horizontal ground. point B is vertically above A , and represents a balloon which is ascending at a steady rate of 3 m s 1 . The balloon is being observed from a moving point Q on the line PA . At time t = 0 s, the balloon is at A and the observer is at P observation point Q moves towards A with steady speed 4 m s 1 . At time t , the angle BQA is radians. Show that d d t 1500 9 t 2 500 4 t 2 .  The . The t.
TJC JC1 Promotional Exams 2011_Solutions and Marker’s Comments 3 Using an algebraic method, solve the inequality 2 2 2 1 x x x .  Hence solve 2 cos 2 cos2 2 2 cos2 1 where 0 2 .  Solution: 2 2 3 2 2 2 2 2 0 1 1 1 x x x x x x x x x x   3 1 0 x x x 0 or 1 3 x x Replacing x by cos2 , cos2 0 or 1 cos2 3 N.A. since 1 cos2 1 Hence, 3 4 2 p p or 5 7 4 4 p p . Marker’s Comments : Some students did cross-multiplication for inequality which was a serious mistake. Most students did not know how to solve inequality involving trigonometrical functions even though GC was allowed.
TJC JC1 Promotional Exams 2011_Solutions and Marker’s Comments 4
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