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2. The last term of a progression is 68, the first term 8, and the number of terms 16. What is the common difference?

Ans. 4. 3. The last term of a progression is 105, the first term 10, and the number of terms 20. What is the common difference ?

Ans. 5.

PROBLEM III. (206.) To find any number of arithmetical means between two given numbers.

In order to solve this problem, we must first find the common difference. The whole number of terms of the series consists of the two extremes and all the intermediate terms. If, then, m represent the required number of means, m+2 will be the whole number of terms.

By substituting m+2 for n in the formula, Art. 205, we obtain

1-a

m+1; that is, the common difference of the required series is equal to the difference between the two given numbers, divided by the number of means plus one.

Having obtained the common difference, the required means are easily obtained by addition. Ex. 1. Find 9 arithmetical means between 2 and 32.

1-a 32-2 d= -

n=10=3, the common difference.

QUEST.-How may we find any number of arithmetical means ba tween two numbers ?

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Hence the progression is

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32. Ex. 2. Find 10 arithmetical means between 3 and 58.

Ans. 8, 13, 18, 23, 28, 33, 38, 43, 48, and 53. Ex. 3. Find 6 arithmetical means between 1 and 50.

Ans. 8, 15, 22, 29, 36, and 43.

(207.)

Examples 1. A student bought 7 books, the prices of which were in arithmetical progression. The price of the cheapest was 5 shillings, and the price of the dearest 23 shillings. What was the price of each book?

Ans. 5, 8, 11, 14, 17, 20, and 23 shillings. 2. What is the nth term of the series 1, 3, 5, 7, 9, etc.?

Ans. 2n-1; that is, the last term of this series is one less than twice the number of terms. 3. What is the sum of n terms of the series 1, 3, 5, 7, 9, etc. ?

. Ans. no; that is, the sum of the terms of this series is equal to the square of the number of terms. Thus, 1+3

= 4=2'. 1+3+5 = 9=3, 1+3+5+7 =16=4,

1+3+5+7+9=25=5'. 4. What is the sum of n terms of the series

Quest.-What is the last term of the series of odd numbers begin. ning with unity? What is the sum of the series of odd numbers be ginning with unity?

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5. What is the sum of n terms of the series 2, 4, 6, 8, 10, etc?

Ans. n(n+1) 6. One hundred stones being placed on the ground in a straight line, at the distance of two yards from each other, how far will a person travel who shall bring them one by one to a basket which is placed two yards from the first stone ?

Ans. 20,200 yards.

GEOMETRICAL PROGRESSION. (208.) A Geometrical Progression is a series of quantities, each of which is equal to the product of that which precedes it by a constant number. Thus the series

2, 4, 8, 16, 32, etc., and

81, 27, 9, 3, etc., are geometrical progressions. In the former, each number is derived from the preceding by multiplying it by 2, and the series forms an increasing geometrical progression. In the latter, each number is derived from the preceding by multiplying it by }, and the series forms a decreasing geometrical progression.

In each of these cases the common multiplier is called the common ratio.

Quest.-What is a Geometrical Progression ? What is an increasing progression? What is a decreasing progression? What is the com mon ratio?

PROBLEM IV. (209.) To find any term of a geometrical progres. sion.

If a represent the first term of the progression, and r the common ratio, the second term of the series will be ar, the third ar*, the fourth ar®, the fifth ar', etc.

The exponent of r in the second term is 1, in the third term is 2, in the fourth term is 3, and so on; hence the exponent of r in the nth term of the series will be n–1; that is, the nth term of the series may be written

- arn-1 If we represent any term of the series by l, and suppose n to be the number which marks the place of that term in the series, the expression for this term will be

l=ar-1. (210.) Hence, if we put l to represent the last term of the series, we shall have the following

RULE. The last term of a geometrical progression is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms.

This rule will enable us to find any term of a series without being obliged to determine all those which precede it.

Quest.—How may we find any term of a geometrical progression ? Give the rule for the last term of a geometrical progression. Explain the reason of the rule.

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Examples. 1. What is the sixth term of a geometrical progres. sion whose first term is 3 and common ratio 2 ?

Ans. l=ar-l=3x2=3x32=96. 2. What is the seventh term of a geometrical progression whose first term is 4 and common ratio 3 ?

. Ans. 2916. 3. What is the eighth term of a progression whose first term is 5 and common ratio 4 ?

Ans. 81920. 4. What is the ninth term of a progression whose first term is 6 and common ratio 3 ?

Ans. 39366. 5. What is the tenth term of a progression whose first term is 7 and common ratio 2 ?

Ans. 3584. 6. What is the ninth term of the series 1, 3, 9, 27, 81, etc. ?

Ans. 6561. 7. What is the eighth term of the series 1, 4, 16, 64, 256, etc.?

Ans. 16384 8. What is the seventh term of the series 1, 5, 25, 125, 625, etc. ?

Ans.. 15625.

PROBLEM V. (211.) To find the sum of the terms of a geomet. rical progression.

If we take any geometrical series, and multiply Quest.—How may we find the sum of the terms of a geometrical progression?

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