Search SchoolNGR

Tuesday, 07 July 2026
Register . Login

(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, ...

(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.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
    Correct Answer: Option
    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

    Share question on: