CHAPTER IV.

Of the Summation of Arithmetical Progressions.

371. It is often necessary also to find the sum of an arithmetical progression. This might be done by adding all the terms together; but as the addition would be very tedious, when the progression consisted of a great number of terms, a rule has been devised, by which the sum may be more readily obtained. 372. We shall first consider a particular given progression, such that the first term = 2, the difference 3, the last term =29, and the number of terms = 10;

8,

11, 14, 17, 20,

23, 26,

29.

2, 5, We see, in this progression, that the sum of the first and the last term = 31; the sum of the second and the last but one 31; the sum of the third and the last but two 31, and so on; and thence we conclude, that the sum of any two terms equally distant, the one from the first, and the other from the last term, is always equal to the sum of the first and the last term.

373. The reasons of this may be easily traced. For, if we suppose the first = a, the last =, and the difference = d, the sum of the first and the last term is a +; and the second term being = a + d, and the last but one = % d, the sum of these two terms is also a + %. Further, the third term being a + 2 d, and the last but two = % - 2 d, it is evident that these two terms also, when added together, make a + z. The demonstration may be easily extended to all the rest.

374. To determine, therefore, the sum of the progression proposed, let us write the same progression term by term, inverted, and add the corresponding terms together, as follows:

2 + 5 + 8 +11 + 14 +17 + 20 + 23 + 26 + 29 29+26+23 + 20 + 17 + 14 + 11 + 8 + 5 + 2. 31 +31 +31 + 31 + 31 + 31 + 31 + 31 + 31 + 31 This series of equal terms is evidently equal to twice the sum of the given progression; now the number of these equal terms is 10, as in the progression, and their sum, consequently, = 10

× 31 310. So that, since this sum is twice the sum of the arithmetical progression, the sum required must be 155.

375. If we proceed in the same manner, with respect to any arithmetical progression, the first term of which is = a, the last =%, and the number of terms = n; writing under the given progression the same progression inverted, and adding term to term, we shall have a series of n terms, each of which will be = a+%; the sum of this series will consequently ben (a + %), and it will be twice the sum of the proposed arithmetical pron (a + gression; which therefore will be = - ✩). 2

376. This result furnishes an easy method of finding the sum of any arithmetical progression; and may be reduced to the following rule:

Multiply the sum of the first and the last term by the number of terms, and half the product will be the sum of the whole progres

sion.

Or, which amounts to the same, multiply the sum of the first and the last term by half the number of terms.

Or, multiply half the sum of the first and the last term by the whole number of terms. Each of these enunciations of the rule will give the sum of the progression.

377. It may be proper to illustrate this rule by some examples.

First, let it be required to find the sum of the progression of the natural numbers, 1, 2, 3, &c. to 100. This will be, by the 100 × 101 first rule,

2

50 × 101 = 5050.

If it were required to tell how many strokes a clock strikes in twelve hours; we must add together the numbers 1, 2, 3, as 12 x 13 far as 12; now this sum is found immediately = 2

=6x

1378. If we wished to know the sum of the same progression continued to 1000, we should find it to be 500500; and the sum of this progression continued to 10000, would be 50005000.

378. Another question. A person buys a horse, on condition that for the first nail he shall pay 5 halfpence, for the second 8, for the third 11, and so on, always increasing 3 halfpence more

for each following one; the horse having 32 nails, it is required to tell how much he will cost the purchaser?

In this question, it is required to find the sum of an arithmetical progression, the first term of which is 5, the difference =3, and the number of terms = 32. We must therefore begin by determining the last term; we find it (by the rule in articles 365 and 370) = 5 + 31 × 3 = 98. After which the sum requir

ed is easily found =

103 X 32
2

105 x 16; whence we conclude

that the horse costs 1648 halfpence, or 3l. 8s. 8d.

d,

379. Generally, let the first term be = a, the difference = and the number of terms = n; and let it be required to find, by means of these data, the sum of the whole progression. As the last term must be = a + (n 1) 4, the sum of the first and last will be = 2a + (n − 1) d. Multiplying this sum by the number of terms n, we have 2 n a + n (n − 1)d; the sum required theren (n - 1) d

fore will be = n a + 2

-

This formula, if applied to the preceding example, or to a =

d = 3, and n = 32, gives 5 × 32 +

32 x 31 x 3
2

1648; the same sum that we obtained before.

: 5, =160 + 1488 =

380. If it be required to add together all the natural numbers from 1 to n, we have, for finding this sum, the first term = 1, the last term = n, and the number of terms = n; wherefore the nn + n n (n + 1) 2

sum required is =

2

If we make n = 1766, the sum of all the numbers, from 1 to 1766, will be = 883 × 1767 = 1560261.

381. Let the progression of uneven numbers be proposed, 1, 3, 5, 7, &c. continued to n terms, and let the sum of it be required: Here the first term is = 1, the difference = 2, the number of terms = n; the last term will therefore be = 1 + (n − 1) 2 = 2n-1, and consequently the sum required = nn.

The whole therefore consists in multiplying the number of terms by itself. So that whatever number of terms of this progression we add together, the sum will be always a square, namely, the square of the number of terms. This we shall exemplify as follows;

Sum,

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, &c. 382. Let the first term be = 1, the difference = 3, and the number of terms = n; we shall have the progression 1, 4, 7, 10, &c. the last term of which will be 1 + ( n - 1) 3 = 3 n 2; wherefore the sum of the first and the last term = 3 n

consequently, the sum of this progression =

n (3 n

2

1, and

1)___ Snn-n

If we suppose n = 20, the sum will be 10 × 59 = 590.

=

2

383. Again, let the first term = 1, the difference = d, and the number of terms = n; then the last term will be = 1 + (n − 1) d. Adding the first, we have 2 + (n— 1) d, and multiplying by the number of terms, we have 2 n + n ( n · 1)d; whence we deduce

the sum of the progression = n +

n(n-1)d

2

We subjoin the following small table :

If d = 1, the sum is = n + n (n − 1) _ nn +n

=

2

CHAPTER V.

Of Geometrical Ratio.

384. THE geometrical ratio of two numbers is found by resolving the question, how many times is one of those numbers greater than the other? This is done by dividing one by the other; and the quotient, therefore, expresses the ratio required.

385. We have here three things to consider; !st, the first of the two given numbers, which is called the antecedent; 2dly, the other number, which is called the consequent; 3dly, the ratio of the two numbers, or the quotient arising from the division of the antecedent by the consequent. For example, if the relation of the numbers 18 and 12 be required, 18 is the antecedent, 12 is the consequent, and the ratio will be = 1; whence we see, that the antecedent contains the consequent once and a half.

18

386. It is usual to represent geometrical relation by two points, placed one above the other, between the antecedent and the consequent. Thus ab means the geometrical relation of these two numbers, or the ratio of b to a.

We have already remarked, that this sign is employed to represent division, and for this reason we make use of it here; because, in order to know the ratio, we must divide a by b. The relation, expressed by this sign, is read simply, a is to b.

387. Relation therefore is expressed by a fraction, whose numerator is the antecedent, and whose denominator is the consequent. Perspicuity requires that this fraction should be always reduced to its lowest terms; which is done, as we have already shewn, by dividing both the numerator and denominator by their greatest common divisor. Thus, the fraction becomes 3, by dividing both terms by 6.

388. So that relations only differ according as their ratios are different; and there are as many different kinds of geometrical relations as we can conceive different ratios.

The first kind is undoubtedly that in which the ratio becomes unity; this case happens when the two numbers are equal, as in 3:3; 4:4; a: a; the ratio is here 1, and for this reason we call it the relation of equality.