Question

X is carrying out a series of experiments which involve using increasing amounts of a chemical. In the first experiment he uses 6g of the chemical and in the second experiment he uses 7.8g of the chemical.

(i) Given that the amounts of the chemical used form an arithmetic progression, find the total amount of chemical used in the first 30 experiments. [4]
(ii) Instead it is given that the amounts of the chemical used form a geometric progression. X has a total of 1800g of the chemical available. Show that N, the greatest number of experiments possible, satisfies the inequality
1.3N ≤ 91.
and use logarithms to calculate the value of N.

Answer 1

a) 963g

b) 17

Explanation:

i) Now we must first find the common difference of the AP

d = 7.8 -6 = 1.8

Now the sum of the AP gives the total amount of chemical used for 30 experiments

Sn = n/2 [2a + (n-1)d]

n= 30, a = 6, d= 1.8

Sn = 30/2[2(6) + (30-1) (1.8)]

Sn = 15[12 + 52.2]

Sn = 963g

ii)For the G.P the common ration is (r)= 7.8/6 = 1.3

Since r>1, sum of GP = a (r^n -1)/r-1

If sum = 1800g

1800= 6(1.3^n -1)/1.3 - 1

1800 = 6(1.3^n -1)/0.3

1800 * 0.3 = 6(1.3^n -1)

540 = 6(1.3^n -1)

540/6 = (1.3^n -1)

90 = 1.3^n -1

90 + 1 = 1.3^n

1.3^n = 91

log 1.3^n = log 91

nlog1.3 = log 91

n = log91/log 1.3

n = 1.959/0.114

n = 17

Answer 2

a) 963g

b) 17

Explanation:

i) Now we must first find the common difference of the AP

d = 7.8 -6 = 1.8

Now the sum of the AP gives the total amount of chemical used for 30 experiments

Sn = n/2 [2a + (n-1)d]

n= 30, a = 6, d= 1.8

Sn = 30/2[2(6) + (30-1) (1.8)]

Sn = 15[12 + 52.2]

Sn = 963g

ii)For the G.P the common ration is (r)= 7.8/6 = 1.3

Since r>1, sum of GP = a (r^n -1)/r-1

If sum = 1800g

1800= 6(1.3^n -1)/1.3 - 1

1800 = 6(1.3^n -1)/0.3

1800 * 0.3 = 6(1.3^n -1)

540 = 6(1.3^n -1)

540/6 = (1.3^n -1)

90 = 1.3^n -1

90 + 1 = 1.3^n

1.3^n = 91

log 1.3^n = log 91

nlog1.3 = log 91

n = log91/log 1.3

n = 1.959/0.114

n = 17