Let SV when T is given,

and S T when V is given;

when neither T nor V is given, S¤ TV.

For the variation of S depends upon the variations of the two quantities T and V; let the changes take place separately, and whilst T is changed to t, let S be changed to S'; then, by the supposition, S:ST:t;

but this value S will again be changed to s, by the variation of V, and in the same proportion that V is changed; that is,

Ex.

If 6 horses can plough 17 acres in 2 days, how many acres will 93 horses plough in 4 days?

The number of acres ploughed ∞ number of horses,

if the days are the same.

∞ number of days,

if the horses are the same.

When both horses and days are different,

number of acres ∞ number of days × number of horses; .. number of acres required: 17 acres :: 93 × 44: 6×2;

number of acres required_93 × 4

279. If there be any number of magnitudes, P, Q, R, S, &c., each of which varies as another, V, when the rest are constant; when they are all changed, V varies as their product.

Let VP when Q, R, and S, are given;

VQ when P, R, and S, are given;
VR when P, Q, and S, are given;

VS when P, Q, and R, are given;

when P, Q, R, S, are all variable, V ∞ PQRS.

Let the changes of V dependent upon the changes of P, Q, R, S take place separately, and whilst P is changed to p, let V be changed to V1; when Qis changed to q, let V, be changed to V; when R is changed to r, let V, be changed to V,; and when S is changed to s, let V, be changed to v. Then, by the supposition, these changes are such, that we have the following proportions:

2

Otherwise,

that is, VPQRS.

VP when Q, R, &c. are invariable,

.. V=mP where m does not contain P. (Art. 269.)

But if P be made constant, and Q be made to vary, VQ, and mPQ; .. also m∞ Q, and consequently m=nQ, where n does not contain Q; and n cannot contain P, by what has been said before, .. V=nPQ. Now if R be considered alone to vary, by the same reasoning we have n∞R, and ... =pR ; .. V=p. PQR, where p does not contain P, Q, or R:

.. V∞ PQR.

This may be extended to any number of magnitudes.

[Exercises Zb.]

280.

ARITHMETICAL PROGRESSION.

DEF. Quantities are said to be in Arithmetical Progression when they increase or decrease by a Common Difference. Thus each of the following series of quantities

Hence it is manifest, that, if a be the first term and a + b the second, a +2b is the third, a + 3b the fourth, &c. and a + (n−1)b the nth term.

By giving the dently of the rest. we have

proper value to n any one term may be found indepenThus, if the 50th term of the first series be required,

a=1, b=2, and n=50,

.. the 50th term =1+(50−1)× 2,

=1+98=99.

281. If the two extremes, and the number of terms, in an Arithmetical Progression be given, the means, that is, the intervening terms, may be found.

Let a and be the extremes, that is, the first and last terms, of an Arithmetical Progression, n the number of terms, and b the unknown common difference; then the progression is

and b being thus found, all the means a+b, a+2b, &c. are known.

Ex. There are four means, or intervening terms, in Arithmetical Progression between 1 and 36; find them.

b-a a+b

COR. A single Arithmetical mean between any two quantities a and b, which is called the Arithmetical mean, will be a+ or- : but this may be shewn more simply thus,

Let a, a, b be in Arithmetical Progression,

282.

then by Def. x-a=b-x; .. x=
x= '1' (a+b).

3-1

2

The sum of a series of quantities in Arithmetical Progression may be found by multiplying the sum of the first and last terms by half the number of terms*.

Let a be the first term, b the common difference†, n the num

ber of terms, the last term, and s the sum of the series:

By this rule we are enabled to find the sum of any number of terms in Arithmetical Progression, without the trouble of adding them all together. ED.

The Common Difference in any proposed case is obviously found by subtracting any term from the term next following. ED.

Then, a+ (a+b) + (a +2b) + ......

also, reversing the order of the terms,

1+ (1−b)+(l−2b) +.........+a=s;

adding, (a+1) + (a + 1) + ( a + 1) + ... ton terms = 28,

COR. 2. Any three of the quantities s, a, n, b, being given, the fourth may be found from the equation

Ex. 1. To find the sum of 14 terms of the series 1, 3, 5, 7, &c.
Here a=1, b=2, n=14;

•. s = (2 +26)×7 = 196.

Ex. 2. Required the sum of 9 terms of the series 11, 9, 7, 5, &c.

In this case a=11, b=-2, n=9;

Ex. 3. If the first term of an arithmetical progression be 14, and the sum of 8 terms be 28, what is the common difference, and the series?

In the case proposed 828, a=14, n=8;

.. b=

56-224 7-28
8x7

3.

7

Hence the series is 14, 11, 8, 5, &c.

Ex. 4. If the first term of an arithmetical progression be 3, the common difference 1, and the sum 22, find the number of terms.

and by solving this quadratic equation it is found that n, the number of terms, is 4.

In this and similar examples the negative value of n is excluded by the supposition under which n originally enters the equation, since a number of terms can only be expressed by a positive integer.

A meaning can however be obtained for the negative value of n. For in the equation s={2a+(n−1).b}1⁄2, write -n for n, and it becomes

or

s={(n+1).b−2a}}, or {2.(b−a)+(n−1).b} 1⁄2 ;

and if we compare this with the original equation, we see that the righthand member is the sum of an arithmetical progression whose first term is b-a, and common difference b: this series therefore is indicated by the negative result.

283. The sum of any two terms taken equidistant from the beginning and end of an arithmetical progression is equal to the sum of the first

and last terms.

This is proved in the demonstration of Art. 282; or it may be shewn independently, thus:-Let a, b, c, d, e, f, g be in arithmetical progression, then, by Definition,

b-a-g-f, or b+f=a+g.

Again, c-b-f-e; . c+e=b+f=a+g. And so on, whatever the number of terms in the series may be.

284. If the number of terms of an arithmetical progression be odd, twice the middle term is equal to the sum of the first and last terms.