Sunday

December 21, 2014

December 21, 2014

Posted by **Anonymous** on Sunday, May 4, 2008 at 11:20am.

(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).

- Maths - Series -
**tchrwill**, Sunday, May 4, 2008 at 1:26pm1(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.

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