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GEOMETRICAL PROGRESSION.

590. Geometrical Progression is a series of numbers increasing or decreasing by a common ratio.

Thus, 2, 6, 18, 54,
and

162, is an ascending series,

64, 32, 16, 8,

4, is a descending series.

In the first series 3 is the ratio, and in the second.

591. In Geometrical Progression there are five elements, of which any three being given the other two can be found:

-:

1. The first term.

2. The last term.

3. The ratio.

4. The number of terms.

5. The sum of all the terins.

592. The first term, ratio, and number of terms given, to find the last term.

In a series, let 2 be the first term and 4 the ratio ;

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In this 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, the index of the power of the ratio always being one less than the number of the term sought. A similar explanation may be given when the series is descending. Hence,

Rule.

Multiply the first term by that power of the ratio whose index is equal to the number of terms less one.

98. The first term of a geometrical series is 5, the ratio 2, and the number of terms 8; what is the last term? Ans. 640.

99. The first term is 9, and the ratio; what is the 6th term? 100. The 1st term is 4, the ratio 1.05; what is the 4th term? 101. What is the amount of $15 at compound interest for 5 years at 6% ?

102. Supposing money at compound interest to double once in 12 years, to what will $100 amount in 60 years?

593. The extremes and number of terms given to find the ratio.

Since the last term is obtained (Art. 592) by multiplying the first term by that power of the ratio whose index is equal to the number of terms less one, hence, conversely,

Rule.

Divide the last term by the first, and of the quotient take that root whose index is one less than the number of terms.

103. The first term in a geometrical series is 3, the last term 768, and the number of terms 5; what is the ratio?

Ans. 4.

104. The extremes are 5 and 625, and the number of terms 4; what is the ratio?

105. The extremes are 8 and 1728, and the number of terms 4; what is the ratio?

594. The extremes and ratio given to find the sum of the series.

Let the series be 2, 10, 50, 250, 1250;

then 2+10+ 50 + 250 +1250

and

Hence,

10+50+250 + 1250 + 6250

6250 2:

-

=

= sum of the series,

=

5 times sum of the series. 4 times the sum of the series.

In subtracting the upper series from the lower all the terms cancel except 2 and 6250, and the remainder will be four times the sum of the series; for once a series from five times a series must leave four

times the series; hence, one fourth of this remainder must be the sum of the given series; but 4 is the ratio less 1. Hence,

Rule.

Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less one.

106. The extremes are 4 and 2916, and the ratio 3; what is the sum of the series? Ans. 4372.

107. The extremes are 3 and 9375, and the ratio 5; what is the sum of the series?

108. What debt will be discharged by 12 monthly payments, the first payment being $1, the second $ 2, and so on in a geometrical series ?

595. The first term, ratio, and number of terms given, to find the sum of the series.

Let the first term be 3, the ratio 2, and the number of terms 6; then the series will be 3, 6, 12, 24, 48, 96.

Let 1, 2, 4, 8, 16, 32, be a series whose first term is 1, but the ratio and number of terms the same as in the first series. The sum of this 64 1 2 1'

But the sum of the

series found by the Rule in Art. 594 is first series is evidently 3 times the sum of this second series; that is,

is

64 1

2 1

× 3; and the 64 is the ratio raised to a power whose index

is equal to the number of terms. Hence,

Rule.

Raise the ratio to a power whose index is equal to the number of terms, subtract one, divide the remainder by the ratio less one, and multiply this quotient by the first term.

109. The first term is 5, the ratio 3, and the number of terms 7; what is the sum of the series?

ALLIGATION.

596. Alligation treats of mixing simple substances of different qualities, producing a compound of some intermediate quality. It is of two kinds, Medial and Alternate.

597. Alligation Medial is the process by which we find the price of the mixture, when the quantities and prices of the simples are given.

110. If a merchant mixes 7 pounds of sugar worth 8 cents a pound, with 6 pounds worth 9 cents, 9 pounds worth 10 cents, and 10 pounds worth 12 cents, what is the value of a pound of the mixture?

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111. A hostler mixes 8 bushels of corn worth 60 cents a bushel with 4 bushels of rye at 75 cents a bushel, 15 bushels of oats at 36 cents a bushel, and 6 bushels of barley at 70 cents a bushel; what is the price a bushel of the mixture?

598. Alligation Alternate is the process of mixing quantities of different values so as to obtain a mixture of a required intermediate value.

NOTE. It is evident that the value of a mixture must be greater than the value of the ingredient of the least value, and less than the value of the ingredient of the greatest value, that is, must be intermediate.

112. A grocer has sugars worth, respectively, 7, 9, 12, and 14 cents a pound. How many pounds of each can he take to make a mixture worth 10 cents?

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Ans. 2 lbs. of 7, 8 of 9, 1 of 12, and 3 of 14.

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If a pound worth 7 is sold for 10 cents, there is a gain of 3 cents (marked +3); and if a pound worth 9 is sold for 10 cents,

a gain of 1 cent; but if a pound worth 12, or 14, is sold for 10, there is a loss of 2, or 4 cents (marked 2 and 4). We take any convenient number of pounds of the 7, 9, and 12, respectively, as 2, 8, and 1, and find a gain of 6 + 8, and a loss of 2, leaving a balance of gain of 14 — 2, or 12, cents. In order to balance this gain of 12 cents we must take as many pounds of the 14 cent as 4 cents (the loss on a pound of the 14 cent) is contained times in 12 cents, that is, 3 pounds. With 2 pounds of the 6, 8 of the 9, 1 of the 12, and 3 of the 14 cents, sold at 10 cents a pound, the gains and losses balance. Hence,

599. To solve an example in alligation alternate,

Rule.

Take such quantities of each ingredient as to make the gains and losses equal.

NOTE 1. If there are but two ingredients the quantities taken must have the same ratio to each other; but if there are more than two ingredients the quantities taken may vary infinitely; that is, there are an infinite number of correct solutions to an example in alligation alternate when there are more than two ingredients. The only restriction is that the gains and losses should be equal.

113. A grocer has 12 pounds of tea worth $0.50 a pound, 8 pounds worth $0.75 a pound which he wishes to mix with teas worth, respectively, $0.62 and $0.70 a pound so as to get a mixture worth $ 0.65 a pound. How many of each of the last two can he take?

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