1(a)Glasses are piled upon each other to form a tower.The rows are numbered,beginning with the top row.There is a single glass in the top row,two glasses in row number 2,3 in 3;etc.This continues to the bottom of the tower.A tower is needed which uses a maximum of 3200 glasses.
(i)How many glasses are required for the first 10 rows? -55
(ii)What is the maximum number of rows that can be built with 3200 glasses?
(iii)How many unused glasses will there be?
(b)(An)n>-1
Sn=A1+A2+.........An=3^n-1
(i)Show that A3=18
(ii)Find An as a function of n.
(iii)Show that the series is a geometric progression.
2(a)Consider the arithmetic progression with terms A3=-2 and A12=23. Find the sum of A1+......+A40 (can do)
(b)In 1800 the population of England was 8 million. The economist Malthus (1766-1834) produced a hypothesis, suggesting:-that the population of england would increase, according to a G.P., by 2% per year
-that the english agriculture production,able to feed 10 million people in 1800,would improve according to an A.P. to feed an extra 400000 people every year
Po represents the english population in 1800 and Pn that population in the year 1800+n:
(i)express,according to Malthus' hypothesis Pn as a function n.
Ao represents the number of people that the english agriculture production can feed in 1800 and An that number in 1800+n:
(ii)express,according to Malthus' hypothesis,An as a function of n
(iii)Calculate the population of england in 1900 and the number of people that the english agriculture production can feed in 1900
(iv)Determine the year from which the english agriculture can no longer feed the english population according to Malthus' hypothesis(-use your calculator by graphing or creating the lists:n=L1;Pn=L2;An=L3 tp compare increases).
1(a)--As each row is added, the sum of glasses in the stack are the triangular numbers defined by T = n(n + 1)/2. Thus, there are T = 10(11)/2 = 55 glasses in the stack of 10 rows.
For 3200 glasses, T = n(n + 1)/2 = 3200 yielding n^2 + n - 6400 = 0. SOlving, n = 79.5156 meaning that 79 rows will use up T = 79(80)/2 = 3160 glasses leaving 40 unused.
Please state only one problem on a post.
To answer the given questions, let's break them down one by one:
1(a)
(i) To find the number of glasses required for the first 10 rows, we can observe that the number of glasses in each row forms an arithmetic sequence with a common difference of 1. So, we can calculate the sum of the first 10 numbers (1+2+3+...+10) using the formula for the sum of an arithmetic series: Sn = (n/2)(A1 + An), where Sn represents the sum, n is the number of terms, A1 is the first term, and An is the last term. Plugging in the values, we get Sn = (10/2)(1 + 10) = 55.
(ii) To find the maximum number of rows that can be built with 3200 glasses, we need to find the value of n (row number) such that the sum Sn of the first n terms is less than or equal to 3200. We can use the formula Sn = (n/2)(A1 + An) and solve for n. Given that A1 = 1 and the sum Sn should be less than or equal to 3200, we can set up the inequality (n/2)(1 + n) ≤ 3200 and solve it. This will give us the maximum number of rows possible.
(iii) To find the number of unused glasses, we can subtract the sum of glasses used for the maximum number of rows (as calculated in part (ii)) from the total number of glasses (3200).
1(b)
(i) To show that A3 = 18, we can use the formula Sn = (n/2)(A1 + An) and plug in the values of n = 3 and Sn = 18 to solve for A1. Then we can calculate A3 by substituting A1 and Sn = 18 in the formula again, giving us A3 = 18.
(ii) To find a general expression for An as a function of n, we can use the formula Sn = (n/2)(A1 + An) and solve for An in terms of Sn and A1.
(iii) To show that the series is a geometric progression, we need to show that the ratio of consecutive terms remains constant. We can calculate the ratios between A2/A1, A3/A2, and so on, and check if they are all equal.
2(a)
To find the sum of A1 + A2 + ... + A40, we can use the formula for the sum of an arithmetic series: Sn = (n/2)(A1 + An) and calculate the sum for n = 40.
2(b)
(i) To express Pn as a function of n, we can start with the initial population in 1800 (Po) and apply the growth rate given by Malthus' hypothesis. Multiplying Po by the growth factor (1.02 for 2% growth rate) for n years will give us Pn.
(ii) To express An as a function of n, we can start with the initial capacity of English agriculture production in 1800 (Ao) and apply the improvement rate given by Malthus' hypothesis. Adding 400000 to Ao for n years will give us An.
(iii) To calculate the population of England in 1900, we can substitute n = 100 into the expression for Pn. Similarly, to find the number of people that English agriculture can feed in 1900, we can substitute n = 100 into the expression for An.
(iv) To determine the year from which English agriculture can no longer feed the English population, we need to compare the population growth (according to Malthus' hypothesis) with the increase in food production. One way to do this is to create a table or graph of Pn and An for various values of n and find the point where the population surpasses the food production capacity.