Examples in Progression.

25. Find the sum of the odd numbers from 1 to 99.

Ans. 2500.

26. Find the sum of the even numbers 2, 4, 6, &c. up

to n terms.

Ans. ́n (n+1).

27. Find the sum of the even numbers from 2 to 100. Ans. 2550.

28. One hundred stones being placed on the ground, in a straight line, at a distance of 2 yards from each other; how far will a person travel, who shall bring them one by one to a basket, placed at 2 yards from the first stone?

Ans. 11 miles, 840 yards.

29. We know, from natural philosophy, that, a body which falls in a vacuum, passes, in the first second of its fall, through a space of 16 feet, but in each succeeding second, 32 feet more than in the immediately preceding one. Now, if a body has been falling 20 seconds, how many feet will it have fallen the last second? and how many in the whole time?

Ans. 627 feet in the last second, and 64333 feet in the whole time.

30. In a foundery, a person saw 15 rows of cannon-balls placed one above another, and asked a bombardier how many balls there were in the lowest row. "You may easily calculate that," answered the bombardier. "In all these rows together, there are 4200 balls, and each row, from the first to the last, contains 20 balls less than the one immediately below it." How many balls, therefore, were there in the lowest row?

Ans. 420.

252. The arithmetical mean between several

Arithmetical Mean.

quantities is the quotient of their sum divided by their number.

Thus the arithmetical mean between the two quantities a and b is half their sum, or (a+b); that between the four quantities 1, 7, 11, 5 is 6.

253. Problem. To find the arithmetical mean between the terms of an arithmetical progression. Scholium. It is, by the preceding definition

that is, half the sum of the extremes, and also, by art. 248, 'half the sum of any two terms at equal distances from the

extremes.

254. Problem. To find the first and last terms of a progression of which the arithmetical mean, the number of terms, and the common difference are known.

Solution. If we denote the arithmetical mean by M, we

have

S
M=

= (a + 1);

n

which, substituted in the result of example 20, in art. 251,

255. Scholium. In very many of the problems involving arithmetical progression, it is convenient

Examples involving Arithmetical Progression.

to use for one of the unknown quantities the arithmetical mean.

256. EXAMPLES.

1. Find five numbers in arithmetical progression whose sum is 25, and whose continued product is 945.

Solution. Denote the arithmetical mean by M, and the common difference by r, and we have, by art. 254,

and the continued product of these terms is

(5—2r)(5—r)5(5+r)(5+2r)=3125-625 r2+20 r2 = 945.

Hence we find

r= ±2, or = ±√ 541⁄2 ;

and the only rational series satisfying the condition is, therefore, 1, 3, 5, 7, 9.

2. Find four numbers in arithmetical progression whose sum is 32, and the sum of whose squares is 276.

Ans. 5, 7, 9, 11.

3. A traveller sets out for a certain place, and travels 1 mile the first day, 2 the second, and so on. In five days afterwards another sets out, and travels 12 miles a day. How long and how far must he travel to overtake the first?

Ans. 3 days and 36 miles.

Examples involving Arithmetical Progression.

4. Find four numbers in arithmetical progression whose sum is 28, and continued product 585.

Ans. 1, 5, 9, 13.

5. The sum of the squares of the first and last of four numbers in arithmetical progression is 200, and the sum of the squares of the second and third is 136; find them.

Ans. 2, 6, 10, 14.

6. Eighteen numbers in arithmetical progression are such, that the sum of the two mean terms is 311, and the product of the extreme terms is 851. Find the first term and the common difference.

Ans. The first term is 3,

the common difference is 11.

SECTION II.

Geometrical Progression.

257. A Geometrical Progression, or a progression by quotients, is a series of terms which increase or decrease by a constant ratio.

a, l, n, and S will be used in this section as in the last, to denote respectively the first term, the last term, the number of terms, and the sum of the terms; and r will be used to denote the constant ratio.

258. Problem. To find the last term of a geometrical progression when its first term, ratio, and number of terms are known.

To find the last Term and Sum.

Solution. In this case a, r, and n are given, to find 7. Now the terms of the series are as follows:

a, ar, ar2, ar3 ... &c. ... arn−1;

so that, the last or nth term is

1=arn-1;

that is, the last term is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms.

259. Problem. To find the sum of a geometrical progression, of which the first term, the ratio, and the number of terms are known.

Solution. We have

S=a+ar+ar2+&c....+arn-2+arn−1.

If we multiply all the terms of this equation by r, we have rs=ar+are+ar3 + &c. ...+arn-1+ar-n; from which, subtracting the former equation, and striking out the terms which cancel, we have

or

rS-Sarn.

a,

whence

(r1) Sar" — a = a (r” — 1) ;

Hence, to find the sum, multiply the first term by the difference between unity and that power of the ratio whose exponent is equal to the number of terms, and divide the product by the difference between unity and the ratio.