Arithmetic Progression question

First term, a = 5kg
Common difference, d = 2kg

Nth day is the nth term

= a + (n -1)d

= 5kg + (n-1)(2kg)
= 5kg + 2n kg - 2kg
= (3 + 2n)kg


But i) should be asking for the total amount of sand since day 1.

So sum of all n days' worth of sand spread

= ( n/2(2a + (n - 1)d) )kg

= ( n/2(2(5) + (n -1)(2)) )kg

= n/2 (10 + 2n - 2)kg

= n/2(8 + 2n)kg

= (n² + 4n) kg


ii)

n² + 4n = 1500

Use the method of completing the square


n² + 4n + (4/2)² = 1500 + (4/2)²

(n + 2)² = 1504

n + 2 = ±√1504

since number of days has to be positive,

n + 2 = √1504

n = √1504 - 2

n ≈ 36.781

Minimum number of days

= 37

Part ii can also be done by the regular quadratic formula and quadratic inequality combination (like what is usually done in O Levels, except O Levels quadratic inequalities are almost always factorisable) or even Guess and check.