Question

An oil company bores a hole 120m deep. Estimate the cost of boring if the cost is k70 for drilling the first meter with an increase in cost of k3per meter for each succeeding meter.

ii. An IPhone 12 pro who’s original Value was K32500 decreases in value by k85 per month. How long will it take before the Iphone’s value falls below k12500?

Answer 1

$i) \space use \space the \space formula$

$S_n=n/2[2a+(120-1)3]=120/2[(2\times70+(357)]=60(140+357)=29820$

ii) let cost in k$^{th}\space month=a+(n-1)d=32500+(k-1)(-85)$

$32500+(k-1)(-85)\lt12500$

$(k-1)(-85)\lt-20000$

$(k-1)(85)\geq20000$

$k-1\geq20000/85$

$k\geq20000/85+1=236.29$

$k=236.29\space months$