OPERATION.

3+19×9=22×9=11x9=99=sum of the series.

2

3+19 will be the half sum of the series, which, being multiplied by 9, the number of terms, gives 99 for the sum of the series.

2. Find the sum of an arithmetical progression, whose first term is 2, last term 100, and number of terms fifty. Ans. 2550.

3. Find the sum of an arithmetical progression, whose first term is 10, common difference 3, and number of terms twenty-one. Ans. 840.

Q. In an arithmetical progression composed of an even number of terms, what is the sum of the extremes equal to? If the number of terms be odd? What is the sum of an arithmetical series equal to ?

GEOMETRICAL PROGRESSION.

236. A GEOMETRICAL PROGRESSION is a series of numbers increasing or decreasing by a constant ratio. Thus, the series

1, 2, 4, 8, 16, 32, 64, &c.

in which each term is formed from the preceding, by multiplying by the number 2, is an increasing geometrical progression;

and the series

64, 32, 16, 8, 4, 2, 1,

in which each term is formed from the preceding by dividing by the number 2, is a decreasing geometrical progression. The number by which we continually multiply or divide is called the common ratio.

2, 6, 18, 54, 162, &c.

is an increasing geometrical progression.

162, 54, 18, 6, 2,

is a decreasing geometrical progression, the common ratio being 3.

Q. What is a geometrical progression? What is an increasing geometrical progression? Decreasing? Give an example of an increasing series. Of a decreasing series. What is the common ratio? In the series 2, 6, 18, &c., what is the common ratio? In the series 54, 18, 6, 2, what is the common ratio? In the series 5, 20, 80, &c.? In the series 100, 20, 4?

237. From the manner in which an increasing geometrical series is formed, we see that

2d term = 1st term multiplied by the common ratio. 3d term = 2d term multiplied by the common ratio, or the 1st term multiplied by the square of the common ratio.

4th term = 3d term multiplied by the common ratio, or the 1st term multiplied by the 3d power of the common ratio.

5th term, in like manner the 1st term multiplied by the 4th power of the common ratio, and so with the other terms.

Hence, we conclude, that any term of an increasing geometrical progression is equal to the first term multiplied by the common ratio raised to a power one less than the number of terms.

--

NOTE. In a decreasing series, we would divide by the common ratio raised to a power less 1 than the number of

terms.

EXAMPLES.

1. Find the sixth term of the progression 1, 3, 9, 27, &c.

OPERATION.

1 × 35=1× 243=243.

1 being the first term, 3 the common ratio, and the number of terms six, the sixth term will be equal to the first term 1, multiplied by the common ratio 3 raised to the fifth power, 5 being 1 less than the number of terms. But 353× 3 × 3 × 3 × 3=243. And 1× 243= 243.

2. Find the seventh term of the geometrical progression, whose first term is 3 and common ratio 2.

Ans. 192.

3. Find the tenth term of the progression 2, 10, 50, 250, &c. Ans. 3906250.

238. In every geometrical progression, the product of the extremes is always equal to the product of any two terms taken equi-distant from the extremes, when the number of terms is even; and to the square of the middle term when the number of terms is odd.

Thus, arranging the terms of a series in an inverse order, 1 2 4 8 16 32

the product of the extremes 1 and 32 is 32, and the same result is found for the product of every pair of equi-distant

terms.

In the following series, which has an odd number of terms, we have

the product of the extremes 2 and 162 is equal to the square of the middle term 18 or 18 x 18.

Q. What is the product of the extremes of a geometrical progression equal to? If the number of terms be odd?

239. In a geometrical progression, the sum of the series is equal to the last term multiplied by the ratio; this product diminished by the first term, and the remainder divided by the ratio less one.

EXAMPLES.

1. What is the sum of the geometrical series whose first term is 2, ratio 3, and number of terms ten.?

OPERATION.

10th term = 2 × 39=39366 (Art. 237).

Sum of the series

39366x3-2-118096=59048.

3-1

2

We must first find the last term (Art. 236), and knowing the last term, we can find the sum of the series by multiply. ing it by the ratio 3; then subtracting the first term 2 from the product, and dividing the remainder by the ratio minus 1. 2. The first term of a geometrical series is 1, the ratio 2, and last term 32; what is the sum of the series? Ans. 63.

3. Find the sum of the first twenty terms of the series 3, 9, 27, 81, &c.

Ans.

Q. What is the sum of a geometrical progression equal to ?

APPENDIX.

MENSURATION;

OR, THE

APPLICATION OF ARITHMETIC TO THE MEASUREMENT OF PLANE AND SOLID BODIES.

240. BEFORE we can apply arithmetic to the measurement of geometrical figures, we must make use of some known magnitude which may serve as a common measure. Such a common measure is called an unit. (Art. 2).

The unit is always of the same kind as the magnitudes which are measured; that is, it must be a line, if lines are measured; a plane, if planes are measured; and a solid, if solids are measured.

AL

Thus, to find the length of the line A B, we make use of a line CD of known length, say 1 inch, 1 foot, or 1 yard, and

apply it to the line CD

A B. It is contained in it six times. The number 6 is then the length of the line A B, and the length will be 6 inches, 6 feet, or 6 yards, according as CD is 1 inch, 1 foot, or 1 yard. CD is the unit for lines, or linear unit.

Again, if the figure to be measured were the square

ABCD, we A

would use a square, U, the side of which

is 1 inch, 1

B

foot, or 1 yard,

&c., as the unit, and apply it to the

squareA BCD.

U

D

C