Ex. 1. The first term of an arithmetical series is 3, the common difference 4, and the number of terms 11; what is the sum of the series?

4 X 10 40; 40+3= 43, the last term.

34346; 46223; and 23 X 11253, Ans.

2. The first term is 5, common difference 8, and number of terms 21; what is the sum of the series? Ans. 1785.

3. The first term is 11⁄2, common difference, and number of terms 41; what is the sum of the series? Ans. 471.

4. A falling body descends 16 feet in the first second of time, and in successive seconds of time the increments of velocity are 32 feet; how far will a body fall in 6 seconds?

Ans. 579 feet.

352. Since the sum of a series is found by multiplying the sum of the extremes by half the number of terms (350), so, conversely, if the sum of a series be divided by half the number of terms, the quotient must be the sum of the extremes, from which, if either extreme be subtracted, the remainder will be the other extreme. Hence,

353.

PROB. 6.—The sum of the series, the number of terms, and either extreme being given, to find the other extreme.

RULE. Divide the sum of the series by half the number of terms; from the quotient subtract the given extreme, and the remainder will be the other extreme.

Ex. 1. The sum of an arithmetical series is 57, the number of terms 6, and the least term 2; what is the greatest term? 6 ÷ 2 = 3; 57319; and 19-217, Ans.

2. The sum of a series is 196, number of terms 7, and least term 7; what is the greatest term? Ans. 49.

3. The sum of a series is 352, number of terms 11, and greatest term 57; what is the least term?

Ans. 7.

4. A falling body descends 10293 feet in 8 seconds, in the 8th second it falls 2414; how far does it fall in the 1st second?

Ans. 16 feet.

5. A gentleman owing 10 creditors $200, paid the 1st $2, and the others in arithmetical progression; what did he pay the last?

Ans. $38.

§ 42.

GEOMETRICAL PROGRESSION.

354. Any series of numbers increasing by a common multiplier or decreasing by a common divisor, is said to be in GEOMETRICAL PROGRESSION; thus,

2, 6, 18, 54, etc., is an ascending series, and

64, 32, 16, 8, etc., is a descending series.

355. Here, as in Arithmetical Progression, the numbers forming the series are called terms; the first and last, extremes; the others, means.

The constant multiplier or divisor is the ratio. The ratio may, in every series, be considered a multiplier, integral when the series is ascending and fractional when it is descending; thus, in the 2d series above, the ratio is 2 if considered as a divisor, and , as a multiplier.

356. Here, also, five particulars claim our attention: .

1st. The first term.

2d. The last term.

3d. The ratio.

4th. The number of terms.

5th. The sum of all the terms.

357. Any three of these five particulars being given, the other two may be found.

358. Twenty cases may arise, but the investigation of sev eral of them requires a knowledge of logarithms and the higher Algebraic equations, and, of the remaining cases, it will be suf ficient for our purpose to present a few.

359. In an ascending series, let 3 be the first term, and 4

3 X4 X 4X4 X 4 = 3 X 4 7685th term.

=

3 X4 X 4X4 X 4X4 3 X 453072 6th term.

Again, in a descending series, let 243 be the first term and

In forming the above series we see that the second term is found by multiplying the first term by the ratio; the third term, by multiplying the first by the square of the ratio; the fourth, by multiplying the first by the cube of the ratio, and so on -the index of the power of the ratio always being one less than the number of the term sought. Hence,

360. PROB. 1.—The first term, ratio and number of terms being given, to find the last or any other assigned term,

RULE. Multiply the first term by that power of the ratio whose index is equal to the number of terms preceding the required term, and the product will be the term sought.

Ex. 1. The first term of a geometrical series is 7, the ratio 3, and the number of terms 5; what is the last term?

5-1=4; 34-81; and 81 × 7 = 567, Ans.

2. The first term of a series is 3, and the ratio 2; what is the ninth term?

9—1—8; 2o 256; and 256 × 3 = 768, Ans.

3. The first term is 64, and the ratio; what is the tenth term? Ans. §.

4. A boy bought 17 oranges, agreeing to pay 1 mill for the 1st, 2 mills for the 2d, and so on in geometrical progression; what was the cost of the 17th orange? Ans. $65.536.

5. What is the amount of $1 at compound interest for 6 years at 6 per cent. per annum? Ans. $1.418519112256.

NOTE. In Ex. 5, the 1st term is $1, the ratio is 1.06, and the number of terms 7.

6. Suppose Gen. Washington had put $100 to interest, Dec. 31, 1780, what, in justice, would be due his heirs, Dec. 31, 1900, allowing it to double every 12 years? Ans. $102400.

7. The estimated value of the estate of the Rosthchilds is now (1855) $40000000; what will be its value in 1975, allowing it to double once in 12 years? Ans. $40960000000.

8. Suppose a farmer to plant 1 kernel of corn and to harvest 1000 kernels, and suppose him to plant his entire crop from year to year and to harvest in the same ratio, what will be the value of his 10th year's crop if 1000 kernels make 1 pint, and he sells his corn at $1.12 per bushel?

Ans. $17578125000000000000000000.

361. Since the last term is obtained (360) by multiplying the first term by that power of the ratio whose index is equal to the number of terms less one, so, conversely,

PROB. 2. The extremes and number of terms being given, to find the ratio,

RULE. Divide the last term by the first, and the quotient will be that power of the ratio whose index is one less than the number of terms; the corresponding root of the quotient will therefore be the ratio.

Ex. 1. The first term in a geometrical series is 3, the last term 192 and the number of terms 4; what is the ratio?

192 ÷ 3 = 64; 4 — 1 = 3; and 3/64 = 4, Ans.

2. The first term is 160, the last term 5, and the number of terms 6; what is the ratio?

51603;6-1 = 5; and 5/31⁄2, Ans. 3. The extremes are 2 and 486, and the number of terms 6; what is the ratio? Ans. 3 or 3.

(a) This rule enables us to find any number of geometrical means between two given numbers; for the number of terms in a series is two greater than the number of means; hence the ratio may be found, and then the series is formed by multiplying the first term by the ratio, by its square, its cube, etc.

4. Find 3 geometrical means between 2 and 512.

5122256; √√256 = √√√256 = √16 = 4, ratio; .. 8, 32, 128 are the means, and 2, 8, 32, 128, 256 = the series. 5. Find the series formed by 1 and 256, and 7 geometrical Ans. 1, 2, 4, 8, 16, 32, 64, 128, 256.

means.

(b) If the same number of means be found between the successive terms of a geometrical series, these means, together with the terms of the original series, will form a new geometrical series.

6. If 3 means be found between the successive terms of the series 3, 768, 196608, what will be the new series thus formed? Ans. 3, 12, 48, 192, 768, 3072, 12288, 49152, 196608. 7. Form a new series by inserting 2 means between the successive terms of the series 128, 16, 2, 4, 3.

Ans. 128, 64, 32, 16, 8, 4, 2, 1, 1, 1, 1, 16, 3.

8

362. Having a geometrical series given, e. g. 3, 12, 48, 192, 768, 3072, 12288, can we devise any short method for ascertaining the sum of all the terms?

Let us multiply each term except the last by the ratio, 4; thus,

Prod. by 4,

3, 12, 48, 192, 768, 3072, [12288], the given series. 12, 48, 192, 768, 3072, 12288;

and we shall evidently form a new series like the old, except the