5. How many years' purchase should be given for a freehold estate, interest being calculated at 3 per cent.?

6. If a perpetual annuity is worth 25 years' purchase, find the amount of an annuity of £625 to continue for 2 years.

7. If a perpetual annuity is worth 20 years' purchase, find the annuity to continue for 3 years which can be purchased for £2522.

8. When the rate of interest is 4 per cent., find what sum must be paid now to receive a freehold estate of £400 a year 10 years hence; having given log 104=20170333, log 6.75565=8296670.

9. Find what sum will amount to £500 in 50 years at 2 per cent., interest being payable every moment; given e-1='3678.

10. If 25 years' purchase must be paid for an annuity to continue n years, and 30 years' purchase for an annuity to continue 2n years, find the rate per cent.

11. A man borrows £5000 at 4 per cent. compound interest; if the principal and interest are to be repaid by 10 equal annual instalments, find the amount of each instalment; having given

log 1.04 0170333 and log 675565=5.829667.

12. A man has a capital of £20000 for which he receives interest at 5 per cent.; if he spends £1800 every year, shew that he will be ruined before the end of the 17th year; having given

log 2=3010300, log 3=4771213, log 7=8450980.

13. The annual rent of an estate is £500; if it is let on a lease of 20 years, calculate the fine to be paid to renew the lease when 7 years have elapsed allowing interest at 6 per cent.; having given

log 106=2.0253059, log4-688385=6710233, log3 118042-4938820.

14. If a, b, c years' purchase must be paid for an annuity to continue n, 2n, 3n years respectively; shew that

a2-ab+b2=ac.

15. What is the present worth of a perpetual annuity of £10 payable at the end of the first year, £20 at the end of the second, £30 at the end of the third, and so on, increasing £10 each year; interest being taken at 5 per cent. per annum?

CHAPTER XIX.

INEQUALITIES.

245. ANY quantity a is said to be greater than another quantity b when a-b is positive; thus 2 is greater than – 3, because 2-(-3), or 5 is positive. Also b is said to be less than a when b -a is negative; thus -5 is less than -2, because -5-(-2), or 3 is negative.

In accordance with this definition, zero must be regarded as greater than any negative quantity.

In the present chapter we shall suppose (unless the contrary is directly stated) that the letters always denote real and positive quantities.

246. If a b, then it is evident that

>

that is, an inequality will still hold after each side has been increased, diminished, multiplied, or divided by the same positive quantity.

which shews that in an inequality any term may be transposed from one side to the other if its sign be changed.

If ab, then evidently b<a;

that is, if the sides of an inequality be transposed, the sign of inequality must be reversed.

If ab, then a-b is positive, and b-a is negative; that is, a-(-b) is negative, and therefore

-a <-b;

hence, if the signs of all the terms of an inequality be changed, the sign of inequality must be reversed.

that is, if the sides of an inequality be multiplied by the same negative quantity, the sign of inequality must be reversed.

249. If ab, and if p, q are positive integers, then

or a bi; and therefore a > b2; that is, a" > b", positive quantity.

ab,

250. The square of every real quantity is positive, and therefore greater than zero.

Thus (ab) is positive;

that is, the arithmetic mean of two positive quantities is greater than their geometric mean.

The inequality becomes an equality when the quantities are equal.

251. The results of the preceding article will be found very useful, especially in the case of inequalities in which the letters are involved symmetrically.

H. H. A.

14

and

Example 1. If a, b, c denote positive quantities, prove that

For

a2+b2+ c2bc+ca + ab;

2 (a3 + b3 + c3) > bc (b+c)+ca (c + a)+ ab (a+b).

whence by addition

It may be noticed that this result is true for any real values of a, b, c.

Again, from (1)

b2-bc + c2bc

.. b3+c3>bc (b+c)

(2);

(3).

By writing down the two similar inequalities and adding, we obtain 2 (a3 + b3+c3)>bc (b+c)+ca (c + a) + ab (a + b).

It should be observed that (3) is obtained from (2) by introducing the factor b+c, and that if this factor be negative the inequality (3) will no longer hold.

Example 2. If x may have any real value find which is the greater, x3+1 or x2+x.

according as x +1 is positive or negative; that is, according as x > or < −1.

If x=1, the inequality becomes an equality.

252. Let a and b be two positive quantities, S their sum and P their product; then from the identity

we have

4ab = (a + b)2 − (a — b)2,

4P = S2 - (a - b), and S2 = 4P+ (a − b)3.

Hence, if S is given, P is greatest when a = b; and if P is given, S is least when

a = b;

that is, if the sum of two positive quantities is given, their product is greatest when they are equal; and if the product of two positive quantities is given, their sum is least when they are equal.

253. To find the greatest value of a product the sum of whose factors is constant.

Let there be n factors a, b, c,

...

k, and suppose that their

sum is constant and equal to s.

...

2

Consider the product abc k, and suppose that a and b are any two unequal factors. If we replace the two unequal factors a+b a+b a, b by the two equal factors the product is increased , 2 while the sum remains unaltered; hence so long as the product contains two unequal factors it can be increased without altering the sum of the factors; therefore the product is greatest when all the factors are equal. In this case the value of each of the n factors is and the greatest value of the product is

n

or

By an extension of the meaning of the terms arithmetic mean and geometric mean this result is usually quoted as follows:

the arithmetic mean of any number of positive quantities is greater than the geometric mean.

Example. Shew that (1"+2′′+3′′+...+n2)"> n2 ([n)" ; where r is any real quantity.