By J. Parry Lewis (auth.)

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**Sample text**

10, 6, 2, -2, .. , -2, .. P. in the form a, a + d, a + 2d, a +3d, ... where a is the first term and d is the common difference, both a and d being allowed to be positive or negative. Clearly the common difference is found by subtracting one term from the following term. In the above examples it is 50, 3, -4 and -1! respectively. When we examine the series in the form a, a +d, a +2d, a +3d, ... P. is given by S=n(a+l) 2 =~{2a+(n-1)d} Proofs of these results are given in Appendix 8. It follows from the first of these that for the positive integers up to n S=1+2+3+4+ ...

17 + 13 +9 + ... , to 6 terms, and to 12 terms 5. 80 + 79·5 + 79 + 78·5 + ... , to 160 terms and to 321 terms EXERCISE 42 ALGEBRA 2. The geometric progression As Malthus pointed out, a population tends to grow in a geometric progression. If two people beget four children, so that two married couples beget a total of eight children, then (provided the sexes work out nicely and that there are no untimely deaths), there will soon be four new pairs of married couples. If this generation shows the same fertility then these four couples will give rise to sixteen children, who, in turn, will give rise to a new generation of thirty-two children.

000 x0·9 , , , , £1,000 in 2 years' time is £810 =£1,000 x 0·9 2 etc. • 44 ALGEBRA If the annuity has a life of 10 years then its present value is S1o= £1,000 (l -0·9 10) l-0·9 log 0·9 = 1·9542 10 x log 0·9 = 10 + 9·542 = l·5420 antilog l·5420 =0·3483 Thus, l - 0·91° = l - 0·3483 = 0·6517 S = £1,000 10 X 0·6517 =£6 517 0·1 ' If the annuity has a life of 20 years then we need to evaluate £1,000 (l-0·9 20 ) l-0·9 which comes to £8,787. For 40 years the present value is £9,853. This example illustrates that if we discount a future income by a rate rn where n is the number of time periods which we have to wait, then income in the remote future has very little value now.