(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres. (i) How many days will it take the jogger to reach a distance of 15km in training? (ii) Calculate the total distance he would have run in the training. (b) The second term of a Geometric Progression (GP) is -3. If its sum to infinity is 25/2, find its common ratios.
Explanation
(a)1) The sequence is an A.P: 500, 750, 1000 .. with a = 500 and d=250; Tn=15000 Using T\(_n\) = a + (n - 1)d 15000 = 500+ (n - 1) x 250 15000 = 500 +250n - 250 250n = 14500 ; n = \(\frac{14500}{250}\) = 58 The jogger will reach a distance of 15km in 58 days. (ii) Finding total distance he would have run in the training Using S\(_n\) =n/2 [2a + (n - 1)d] n= 58, d=250, a = 500 = 58/2 [2 x 500 + (58 -1) x 250] = 29[ 1000 +(57 x 250)] = 29 (1000+ 14250), = 29 x 15250 = 442250 (b) 2nd term => ar = -3; a= -3/r S ∞ = \(\frac{a}{1-r}\) = 25/2 \(\frac{-3}{r}\) x \(\frac{1}{1-r}\) = 25/2; 25r - 25r\(^2\) = -6; 25r\(^2\) - 25r - 6 = 0 (5r +1) (5r - 6) = 0 r= -1/5 and 6/5
