(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