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2. If the first term be 4, the common difference 3, and the number of terms 100, what is the last term ?
Ans. 301. 3. There are, in a certain triangular field, 41 rows of corn ; the first row, in 1 corner, is a single hill, the second contains 3 hills, and so on, with a common difference of 2; what is the number of hills in the last row?
A. 81 hills. 4. A man puts out $1 at 6 per cent. simple interest, which, in 1 year, amounts to $1,06 in 2 years to $1,12, and so on, in arithmetical progression, with a common difference of $0 ,06 ; what would be the amount in 40 years?
A. $3,40. Hence we see, that the yearly amounts of any sum, at simple interest, form an arithmetical series, of which the principal is the first term, the last amount is the last term, the yearly interest is the common difference, and the num. ber of years is 1 less than the number of terms.
It is often necessary to find the sum of all the terms, in an arithmetical progression. The most natural mode of obtaining the amount would be to add them together, but an easier method may be discovered, by attending to the following explanation.
1. Suppose we are required to find the sum of all the terms, in a series, whose first term is 2, the number of terms 10, and the common difference 2.
4, 6, 8, 10, 12, 14, 16, 18, 20 20, 18, 16, 14, 12, 10, 8, 6, 4, 2 22, 22, 22, 22, 22, 22, 22, 22, 22, 22,
The first row of figures above, represents the given series. The second, the same series with the order inverted, and the third, the sums of the additions of the corresponding terms in the two series. Examining these series, we shall find that the sums of the corresponding terms are the same, and that each of them is equal to the sum of the extremes, viz. 22. Now as there are 10 of these pairs in the two series, the sum of the terms in both, must be 22X 10=220.
But it is evident, that the sum of the terms in one series, can be only half as great as the sum of both, therefore, if
we divide 220 by 2, we shall find the sum of the terms in one series, which was the thing required. 220-_2=110, the sum of the given series.
From this illustration we derive the following rule';
When the extremes and number of terms are given, to find the sum of the terms,
Multiply the sum of the extremes by the number of terms, and divide the product by 2.
2. The first term of a series is 1, the last term 29, and the number of terms 14. What is the sum of the series ?
A. 210. 3. 1st term, 2, last term, 51, number of terms, 18. Re. quired the sum of the series.
A. 477. 4. Find the sum of the natural terms 1, 2, 3, &c. to 10,000.
A. 50,005,000. 5. A man rents a house for $50, annually, to be paid at the close of each year; what will the rent amount to in 20 years, allowing 6 per cent., simple interest, for the use of the money?
The last year's rent will evidently be $50 without in. terest, the last but one will be the amount of $50 for 1 year, the last but two the amount of $50 for 2 years, and so on, in arithmetical series, to the first, which will be the amount of $50 for 19 years : $107.
If the first term be 50, the last term 107, and the number of terms 20, what is the sum of the series ? A. 1570.
6. What is the amount of an annual pension of $100, being in arrears, that is, remaining unpaid, for 40 years, allowing 5 per cent. simple interest?
A. $7900. 7. There are, in a certain triangular field, 41 rows of corn; the first row, being in 1 corner, is a single hill, and the last row, on the side opposite, contains 81 hills; how many hills of corn in the field ?
A. 1681. The method of finding the common difference, may be learned by what follows.
1. A man bought 100 yards of cloth in Arithmetical progression : for the first yard he gave 4 cents, and for the last 301 cents, what is the common increase on the price of each yard ?
As he bought 100 yards, and at an increased price upon every yard, it is evident that this increase was made 99
times, or once less than the 'number of terms in the series. Hence the price of the last yard was greater than the first, by the addition of 99 times the regular increase.
Therefore if the first price be subtracted from the last, and the remainder be divided by the number of additions (99), the quotient will be the common increase ; 301–4= 297 and 297-:-99=3, the common difference.
Hence, when the extremes and number of terms are given, to find the common difference,
Divide the difference of the extremes, by the number of terms less 1.
2. Extremes 3 and 19; number of terms 9. Required the common difference.
A. 2. 3. Extremes 4 and 56 ; number of terms 14. Required the common difference.
A. 4. 4. A man had 15 houses, increasing equally in value, from the first, worth $700, to the 15th, worth $3500. What was the difference in value between the first and second ?
A. 200. In Arithmetical progression, any three of the following terms being given, the other two may be found. 1. The first term. 2. The last term. 3. The number of terms. 4. The common difference. 5. The sum of all the terms.
GEOMETRICAL PROGRESSION. Any series of numbers, consisting of more than two terms, which increases by a common multiplier, or decrea. ses by a common divisor, is called a Geometrical Series.
Thus the series 2, 4, 8, 16, 32, &c. consists of terms, each of which is twice the preceding, and this is an increasing or ascending Geometrical series.
The series 32, 16, 8, 4, 2, consists of numbers, each of which is one half the preceding, and this is a decreasing or descending Geometrical series.
The common multiplier or divisor is called the Ratio, and the numbers which form the series are called Terms.
As in Arithmetical, so in Geometrical progression, if In arithmetical progression, what are the terms used, and what are the rules for finding them ?
any three of the five following terms be given, the other two may be found.
1. The first term. 2. The last term. 3. The number of terms. 4. The common difference. 5. The sum of all the terms.
1. A man bought a piece of cloth containing 12 yards, the first yard cost 3 cents, the second 6, the third 12, and so on, doubling the price to the last, what cost the last yard? 3x2x2x2x2x2x2x2x2x2x2x2=3x211=6144 Ans.
In examining the above process, it will be seen, that the price of the second yard is found by multiplying the first payment into the ratio (2) once ; the price of the third yard, by multiplying by 2 twice, &c., and that the ratio (2) is used as a factor eleven times, or once less than the num. ber of terms. The last term then, is the eleventh power of the ratio (2) multiplied by the first term (3).
Hence the first term, ratio, and number of terms, being given, to find the last term.
Multiply the first term, by that power of the ratio, whose index is one less than the number of terms.
Note. In involving the ratio, it is not always necessary to produce all the intermediate powers; the process may often be abridged, by multiplying together two powers already obtained, thus, The 11th power the 6th power X the 5th power,
&c. 2. If the first term is 2, the ratio 2, and the number of terms 13, what is the last term ?
A. 8,192. 3. Find the 12th term of a series, whose first term is 3, and ratio, 3.
A. 531,441. 4. A man plants 4 kernels of corn, which, at harvest, produce 32 kernels; these he plants the second year; now, supposing the annual increase to continue 8 fold, what would be the produce of the 16th year, allowing 1000 kernels to a pint?
A. 2199023255,552 bushels. 5. Suppose a man had put out one cent at compound interest in 1620, what would have been the amount in 1824, allowing it to double once in 12 years ? 217=131072.
When are numbers in geometrical progression ? What are the terms used? What are the rules for finding them
The most obvious method of obtaining the sum of the terms in a Geometrical series, might be by addition, but this is not the most expeditious, as will be seen.
1. A man bought 5 yards of cloth, giving 2 cents for the first, 6 cents for the second, and so in 3 fold ratio ; what did the whole cost him?
2, 6, 18, 54, 162
6, 18, 54, 162, 486 The first of the above lines, represents the original series. The second, that series, multiplied by the ratio 3.
Examining these series, it will be seen that their terms are all alike excepting two : viz. the first term of the first series, and the last of the second series. If now we subtract the first series from the last, we have for a remainder 486—2=484, as all the intermediate terms vanish in the subtraction.
Now the last series is three times the first, (for it was made by multiplying the first series by 3,) and as we have already subtracted once the first, the remainder must of course be twice the first.
Therefore if we divide 484 by 2, we shall obtain the sum of the first series.
584-;-2=242 Ans. As in the preceding process, all the terms vanish in the subtraction, excepting the first and last, it will be seen, that the result would have been the same, if the last term only, had been multiplied, and the first subtracted from the product.
Hence, the extremes and ratio being given, to find the sum of all the terms,
Multiply the greater term by the ratio, from the product subtract the least term, and divide the remainder by the ratio less 1.
2. Given the first term, 1 ; the last term, 2,187 ; and the ratio, 3 ; required the sum of the series. A. 3,280.
3. Extremes, i and 65,536; ratio 4 ; required the sum of the series. A. 87,381. 4. Extremes, 1,024 and 59,049; required as above.
A. 175,099. 5. What is the sum of the series 16, 4, 1, 4, ja, og and so on, to an infinite extent?
A. 21). Here it is evident, the last term is 0, or indefinitely near