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771. To find the Number of Terms, when the Extremes and the Common Difference are given.

1. The extremes of an arithmetical series are 4 and 37, and the common difference 3; what is the number of terms?

ANALYSIS. The last term of a series is equal to the first term increased or diminished by the product of the common difference by the number of terms less 1. (Art. 769.)

Now 37-4, or 33, is the product of the common difference 3, by the number of terms less 1. Consequently 33÷3, or 11, must be the number of terms less 1; and 11 + 1, or 12, is the answer required. Hence, the

RULE.-Divide the difference of the extremes by the common difference, and add 1 to the quotient.

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2. The age of the youngest child of a family is 1 year, the oldest 22, and the common difference of their ages 3 yr.; how many children in the family?

3. The extremes of an arithmetical series are 8 and 96, the common difference 4; what is the number of terms?

4. A laborer worked for 50 cts. the first day, 54 cts. the second, 58 cts. the third, and so on till his wages were $2 a day; how many days did he work?

772. To find the Common Difference, when the Extremes and the Number of Terms are given.

1. The extremes of a series are 3 and 21, and the number of terms is 10; what is the common difference?

ANALYSIS. The difference of the extremes 21-8-18, is the product of the number of terms less 1 by the common difference, and 10-1, or 9, is the number of terms less 1; therefore 18÷9, or 2, is the common difference required. (Art. 764.) Hence, the

RULE.-Divide the difference of the extremes by the number of terms less 1.

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2. The ages of 10 children form an arithmetical series; the youngest is 3 yr. and the eldest 30 years; what is the difference of their ages?

3. A military company appropriated $108 for 8 target prizes, the highest of which was $24, and the lowest $34 what was the common difference in the prizes?

4. The amount of $600 for 45 yr. at simple interest is $3120; what is the rate per cent?

5. The amount of $1500 for 27 years is $1620; what is the rate per cent?

773. To find the Sum of all the terms, when the Extremes and the Number of Terms are given.

1. Required the sum of the series having 7 terms, the extremes being 3 and 15.

ANALYSIS.—(1.) The series is 3, 5, 7, 9, 11, 13, 15.

(2.) Inverting the same,

(3.) Adding (1.) and (2.),

(4.) Dividing (3.) by 2,

15,
18+18 + 18 + 18 + 18 +18+18
9+9+9+ 9 + 9 + 9 + 9

13, 11, 9, 7, 5, 3.

twice the sum. 63, the sum.

By inspecting these series, we discover that half the sum of the extremes is equal to the average value of the terms. Hence, the

RULE.-Multiply half the sum of the extremes by the

number of terms.

FORMULA.

= 8

Sa + l
2

x n.

NOTE. From the preceding illustration we see that,

The sum of the extremes is equal to the sum of any two terms equidistant from them; or, to twice the sum of the middle term, if the number of terms be odd.

2. How many strokes does a

12 hours?

common clock strike in

3. Find the sum of all the terms, the extremes being 0 and 300, and the number of terms 1200.

4. A father deposited $1 in the bank for his daughte

her first birthday, $4 the next, $7 the next, and s

much did she have when she was 21 years old?

GEOMETRICAL PROGRESSION.

DEFINITIONS.

774. A Geometrical Progression is a series of numbers which increase or decrease by a common ratio.

775. The Terms of a geometrical progression are the numbers which form the series.

NOTE. The series is called Ascending or Descending, according as the terms increase or decrease. (Arts. 763, 764.)

776. In an ascending series the ratio is greater than one. Thus, 2, 4, 8, 16, 32, 64, etc., is an ascending progression.

777. In a descending series the ratio is less than one.

Thus, 1,,,,,, etc., is a descending progression.

778. In Geometrical Progression there are also five elements or parts to be considered, viz.: the first term, the last term, the number of terms, the ratio, and the sum of all the terms.

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779. To find the Last Term, when the First Term, the Ratio, and the Number of Terms are given.

1. Required the last term of an ascending series having 6 terms, the first term being 3, and the ratio 2.

ANALYSIS. From the definition, the series is

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3× (2 × 2 × 2 × 2), etc. Or,
3 × 24, etc.

Now, 3x 253 × 32:

96, Ans. That is,

Each successive term =

1st term x ratio raised to a power whose ex

ponent is one less than the number of the term. Hence, the

RULE.—Multiply the first term by that power of the ratio whose exponent is 1 less than the number of terms. FORMULA.-l = a × pm – 1 ̧

NOTES.-1. Any term in a series may be found by the preceding rule. For, the series may be supposed to stop at that term.

2. The preceding rule is applicable to Compound Interest; the principal being the first term of the series; the amount of $1 for 1 year the ratio; and the number of years plus 1, the number of terms.

2. A father promised his son 1 ct. for the first example he solved, 2 cts. for the second, 4 cts. for the third, etc.; what would the son receive for the tenth example?

3. What is the amt. of $375 for 4 yr., at 5% compound int.? 4. What is the amount of $1200 for 5 years, at 6% compound interest? Of $2500 for 4 years, at 7% ?

780. To find the First Term, when the Last Term, the Ratio, and the Number of Terms are given.

1. The last term of a progression is 96, the number of terms. 6, and the ratio 2; what is the first term?

ANALYSIS.-Reversing the steps of the preceding rule, we have 96÷25 = 96÷32 = 3, Ans. Hence, the

=

RULE.--Divide the last term by that power of the ratio whose exponent is 1 less than the number of terms.

FORMULA.-a = 1 ÷ 2-1.

2. The last term of a series is 192, the ratio 3, and the number of terms 7; what is the first term?

781. To find the Sum of all the Terms, when the Extremes and Ratio are given.

1. Required the sum of the series whose first and last terms are 2 and 162, and the ratio 3.

ANALYSIS.—Since each succeeding term is found by multiplying the preceding term by the ratio, the series is 2, 6, 18, 54, 162.

(1.) The sum of the series, =2+6+18+54 +162.

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RULE.-Multiply the last term by the ratio, and subtracting the first term from the product, divide the remainder by the ratio less 1.

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2. The first term is 4, the ratio 3, and the last term is 972; what is the sum of the terms?

3. What sum can be paid by 8 instalments; the first being $1, the second $2, etc., in a geometrical series?

4. A man bought a dozen sheep, agreeing to pay 1 ct. for the first, 2 cts. for the second, 4 cts. for the third, etc.; what did he pay for the 12 sheep?

5. A housekeeper bought 12 chairs, paying 2 cts. for the first, 6 cts. for the second, and so on; what did they cost?

782. To find the Sum of a Descending Infinite Series, when the First Term and Ratio are given.

NOTE.-In a descending infinite series the last term being infinitely small, is regarded as 0. Hence, the

RULE.-Divide the first term by the difference between the ratio and 1, and the quotient will be the sum required.

1. What is the sum of the series, 1, 1, 1, continued to infinity, the ratio being ? Ans. 13.

NOTE. The preceding problems in the Progressions embrace their ordi. nary applications. Others might be given, but they involve principles with which the pupil is not supposed to be acquainted.

QUESTIONS.

760. What is progression? 761. The terms? 763. An arithmetical progression? 765. How is each term found in an ascending series? 766. In a descending series?

767. Name the parts. 768. How find the last term? 770. The first term? 771. Number of terms. 772. The common difference? 773. The sum of all the terms?

774. What is geometrical progression? 778. Name the parts. 779. How find the last term? 780. The first? 781. The sum of all the terms? 782. The sum of a descending infinite series?

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