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3. A person traveling, goes' 5 miles the first day, 10 miles the second day, 20 miles the third day, and so on, increasing in geometrical progression. If he continue to travel in this way for 7 days, how far will he go the last day?

CASE II. If we multiply all the terms of a geometrical progression by the ratio, we shall obtain a new progression, whose first term equals the second term of the old progression; the second term of our new progression will equal the third term of the old progression, and so on, for the succeeding terms. Hence, the sum of the old progression, omitting the first term, equals the sum of the new progression, omitting its last term. The sum of the new progression is equal to the old progression, repeated as many times as there are units in the ratio. Therefore, the difference between the new progression and the old progression, is equal to the old progression repeated as many times as there are units in the ratio, less one. But we also know, that the difference between these progressions is equal to the last term of the new progression, diminished by the first term of the old progression; and since the new progression was formed by multiplying the respective terms of the old progression by the ratio, it follows that the last term of the new progression is equal to the last term of the old progression, repeated as many times as there are units in the ratio. Therefore, the last term of the new progression, diminished by the first term of the old progression, is equal to the last term of the old progression, repeated as many times as there are units in the ratio, diminished by the first term of the old progression. Hence, we finally obtain this condition :

That the sum of all the terms of a geometrical progression, repeated as many times as there are units in the ratio, less one, is equal to the last term, multiplied by the ratio, and diminished by the first term. .

Hence, when we have given the first term of a geometrical progression, the last term, and the ratio, to find the sum of all the terms, we have this

RULE... Subtract the first term from the product of the last term into the ratio ; divide the remainder by the ratio, less one.

*Examples. 1. The first term of a geometrical progression is 4, the last term is 78732, and the ratio is 3. What is the sum of all the terms?

In this example the first term subtracted from the product of the last term, into the ratio, is 236192, which divided by the ratio, less one, gives 118096, for the sum of all the terms.

2. The first term of a geometrical progression is 5, the last term is 327680, and the ratio is 4. What is the sum of all the terms?

Ans. 436905. 3. A person sowed a peck of wheat, and used the whole crop for seed the following year; the produce of this 2d year again for seed the 3d year, and so on. If, in the last year his crop is 1048576 pecks, how many pecks did he raise in all, allowing the increase to have been in a four-fold ratio. :

CASE III. Since by case I. the last term is equal to the first term, multiplied into a power of the ratio, whose exponent is equal to the number of terms, less one, it follows that the first term is equal to the last term, divided by the power of the ratio, whose exponent is one less than the number of terms.

Hence, when we have given the last term, the ratio, and the number of terms, to find the first term, we have this

RULE. Divide the last term by a power of the ratio, whose exponent is one less than the number of terms.

Examples. 1. The last term of a geometrical progression is 1048576, the ratio is 4, and the number of terms is 11. What is the first term ?,

In this example the ratio 4 raised to a power, whose index is 10, one less than the number of terms, is 410=1048576, .. 1048576, divided by 1048576, gives 1, for the first term.

2. A man has 6 sons, among whom he divides his estate in a geometrical progression, whose ratio is 2; the last son re. ceived $4800. How much did the first son receive ?

... Ans. $150. 3. A person bought 10 bushels of wheat, paying for it in geometrical progression, whose ratio is 3; the last bushel cost him $196.83. What did he give for the first bushel ?

CASE IV. We also discover from case I. that the last term divided by the first term, will give the power of the ratio, whose exponent is the number of terms, less one.

Hence, when we have given the first term, the last term, and the number of terms, to find the ratio, we have this

RULE. Divide the last term by the first term, extract that root of the quotient which is denoted by the number of terms, less one.

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. Examples.. . 1. The first term of a geometrical progression is 1, the last term is 64, and the number of terms is 7. What is the ratio?

In this example the last term divided by the first term is 64, the number of terms, less one, is 6, .'. we must extract the 6th root of 64 ; we first extract the square root, which is 8, we now extract the cube root of 8, which is 2, for the ratio.

2. In a country, during peace, the population increased every year in the same ratio, and so fast, that in the space of 5 years it became from 10000 to 14641 souls. By what ratio was the increase, yearly?

Ans. 14. .. 3. The first term of a geometrical progression is 4, the last term is 78732, and the number of terms is 10. What is the ratio ?

. . CASE V. . . : If, in case II. we write the product of the first term, into the power of the ratio, whose exponent is the number of terms, less one, instead of the last term, as drawn from case I. we shall have the sum of all the terms, repeated as many times as there are units in the number of terms, less one, equal to the power of the ratio whose exponent is equal to the number of terms, diminished by one and multiplied by the first term.

Hence, when we have given the first term, the ratio, and the number of terms, to find the sum of all the terms, we have this

RULE. From the power of the ratio, whose exponent is the number of terms, subtract one, divide the remainder by the ratio, less one, and multiply the quotient by the first term.

Examples. 1. The first term of a geometrical progression is 3, the ratio is 4, and the number of terms 9. What is the sum of all the terms ?

In this example the ratio raised to a power whose exponent is the number of terms, is 49=262144: this, diminished by one, becomes 262143, which, divided by 3, gives 87381; this, multiplied by the first term, becomes 87381 X3=262143, for the sum of all the terms.

2. A king in India, named Sheran, wished, according to the Arabic author Asephad, that Sessa, the inventor of chess, should himself choose a reward. He requested the grains of wheat, which arise, when 1 is calculated for the first square of the board, 2 for the second square, 4 for the third, and so on; reckoning for each of the 64 squares of the board. twice as many grains as for the preceding. When it was calcula. ted, to the astonishment of the king, it was found to be an enormous number. What was it?

Ans. 18446744073709551615 ; , a sum which, according to a moderate calculation, could not be obtained from the whole earth, in upwards of 70 years, if all the land were employed in the cultivation of wheat.

3. A gentleman married his daughter on New Year's day, and gave her husband 1 shilling towards her portion, and was to double it on the first day of every month during the year. What was her portion ?

CASE VI. We know from case V. that the sum of all the terms multi. plied by the ratio, less one, is equal to one subtracted from the power of the ratio, whose exponent is the number of terms, and this remainder multiplied by the first term.

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