Examples.

1. What debt could be discharged in a year, by weekly payments in arithmetical progression, the first payment being $5, and the last $100?

2. A person agreed to build 56 rods of fence; for the first rod he was to receive 6 cents, for the second, 10 cents, and so on: what did he receive for the last rod, and how much for the whole?

3. If a person travels 30 miles the first day, and a quarter of a mile less each succeeding day, how far will he travel in 30 days?

4. If 120 stones be laid in a straight line, each at a distance of a yard and a quarter from the one next to it, how far must a person travel who picks them up singly and places them in a heap, at the distance of 6 yards from the end of the line and in its continuation?

CASE IV.

398. Having given the first and last terms, and the common difference, to find the number of terms.

1. The first term of an arithmetical progression is 5, the common difference 4, and the last term 41: what is the number of terms?

41

OPERATION.

5 = 36 4)36( = 9 9110 No. terms

ANAJ YSIS. Since the last term is equal to the first teria added to the product of the common difference, by one less than the number of terms (Art. 395), it follows that, if the first term be taken from the last term, the difference will be equal to the product of the common difference by 1 less than the number of terms: if this be divided by the common difference, the quotient will be 1 less than the number of terms.

Rule.-Divide the difference of the two extremes by the common difference, and add 1 to the quotient: the sum will be the number of terms.

Examples.

1. A farmer sold a number of bushels of wheat; it was agreed that, for the first bushel, he should receive 50 cents, and an increase of 9 cents for each succeeding bushel, and for the last, he received $500: how many bushels did he sell?

2. A person proposes to make a journey, and to travel 15 miles the first day, and 33 miles the last, with a daily increase of 14 miles in how many days did he make the journey, and what was the whole distance traveled?

:

3. I owe a debt of $2325, and wish to pay it in equal installments, the first payment to be $575, the second, $500, and decreasing by a common difference, until the last payment which is $200 what will be the number of installments?

GEOMETRICAL PROGRESSION.

399. A GEOMETRICAL PROGRESSION is a series of terms, each of which is derived from the preceding one, by multiplying it by a constant number. The constant multiplier, is called the ratio of the progression.

400. AN INCREASING SERIES is one whose ratio is greater than 1:

A DECREASING SERIES is one whose ratio is less than 1 Thus,

1 2, 4, 8, 16, 32, &c.-ratio 2-increasing series: 32, 16, 8, 4, 2, 1, &c.-ratio decreasing series.

The several numbers resulting from the multiplication, are called terms of the progression. The first and last terms are

called the extremes, and the intermediate terms are called

means.

401. In every Geometrical, as well as in every Arithmetical Progression, there are five parts:

1st, The first term;

2d, The last term;
3d, The common ratio;

4th, The number of terms;

5th, The sum of all the terms;

If any three of these parts are known, or giver, the remain. ing ones can be determined.

CASE I.

402. Having given the first term, the ratio, and the number of terms. to find the last term.

1. The first term is 4, and the common ratio 3: what is the 5th term?

multiplications being 1 less than the number of terms: thus,

4 =

4, 1st term,

3 X 4 = 12, 2d term,

3 × 3 × 4 36, 3d term.

3 × 3 × 3 × 4 = 108, 4th term.

3 × 3 × 3 × 3 × 4 = 324, 5th term.

Therefore, the last term is equal to the first term multiplied by the ratio raised to a power whose exponent is 1 less thar the number of terms.

Rule.-Raise the ratio to a power whose exponent is 1 less than the number of terms, and then multiply this power by the first term.

Examples.

1. The first term of a decreasing progression is 2187; the ratio is, and the number of terms 8: what is the last term?

2. The first term of an increasing geometrical series is 8, the ratio 5: what is the 9th term?

3. The first term of a decreasing geometrical series is 729, the ratio what is the 10th term?

4. If a farmer should sell 15 bushels of wheat, at 1 mill for the first bushel, 1 cent for the second, 1 dime for the third, and so on; what would he receive for the last bushel?

5. A man dying left 5 sons, and bequeathed his estate in the following manner: to his executors, $100; to his youngest son twice as much as to the executors, and to each son double the amount of the next younger brother: what was the eldest son's portion?

6. A merchant engaging in business, trebled his capital once In 4 years if he commenced with $2000, what would his capital amount to at the end of the 12th year?

7. A farmer wishing to buy 16 oxen of a drover, finally agreed to give him for the whole the cost of the last ox only. He was to pay 1 cent for the first, 2 cents for the second, and doubling on each one to the last how much would they cost him?

8. What is the amount of $500 for 3 years at 6 per cent. compound interest?

NOTE.-The ratio is 1.06.

CASE II.

403. Knowing the two extremes and the ratio, to find the sum of the terms.

1. What is the sum of the terms of the progression 2, 6, 18, 54, 162?

OPERATION.

61854 + 162 + 486 = 3 times.

2+6+18 +54 + 162

= 1 time.

ANALYSIS.-If we multiply the terms of the progression by the ratio 3, we have a second progression, 6, 18, 54, 162, 486, which is 3 times as great as the first. If from this we subtract the first, the remainder, 486-2, will be 2 times as great as the first; and if this remainder be divided by 2, the quotient will be the sum of the terms of the first progression.

But 486 is the product of the last term of the given progression multiplied by the ratio; 2 is the first term; and the divisor 2, 1 less than the ratio: hence,

Rule.-Multiply the last term by the ratio; take the dif ference between this product and the first term, and divide the remainder by the difference between 1 and the ratio.

NOTE. When the progression is increasing, the first term is subtracted from the product of the last term by the ratio, and the divisor is found by subtracting 1 from the ratio. When the progression is decreasing, the product of the last term by the ratio is subtracted from the first term, and the ratio is subtracted from 1.

Examples.

1. The first term of a progression is 4, the ratio 3, and the last term 78722: what is the sum of the terms?

2. The first term of a progression is 1024, the ratio, and the last term 4: what is the sum of the series?

3. What debt can be discharged in one year by monthly payments, the first being $2, the second $8, and so on to the end of the year; and what will be the last payment?

4. A gentleman being importuned to sell a fine horse, said