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hcpscyruscm
01.01.2021 •
Mathematics
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.
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Ответ:
a) 963g
b) 17
Step-by-step 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
Ответ: