6. A man bought 10 bushels of wheat on the condition that he should pay 1 cent for the first bushel, 3 for the second, 9 for the third, and so on to the last: what did he pay for the last bushel, and for the ten bushels?

Ans. Last bushel, $196 83; total cost, $295, 24. 189. When the progression is decreasing, we have r<1 and <a; the above formula for the sum is then written under the form

in order that both terms of the fraction may be positive. By substituting arm-1 for 7, in the expression for S,

1. Find the sum of the first five terms of the progression

2. Find the sum of the first twelve terms of the progression

We perceive that the principal difficulty consists in obtaining the numerical value of the last term, a tedious operation, even when the number of terms is not very great.

we make r == 1, it reduces to

S=

This result sometimes indicates indetermination; but it often arises from the existence of a common factor in both numerator and denominator of the fraction, which factor becomes 0, in consequence of a particular supposition.

Such is the fact in the present case, since both terms of the fraction contain the factor r1, which becomes 0, for the particular supposition r = = 1.

If we divide both terms of the fraction by this common factor, we shall find (Art. 60),

We ought to have obtained this result; for, under the supposition made, each term of the progression became equal to a, and since there are n of them, their sum should be na.

furnishes the means for resolving the following problem, viz.

To find m geometrical means between two given numbers a and b; that is, to find a number m of means, which will form with a and b, considered as extremes, a geometrical progression.

To find this series, it is only necessary to know the ratio. Now, the required number of means being m, the total number of terms considered, will be equal to m +2.

have b; therefore, the value of r becomes

=

Moreover, we

To find the ratio, divide the second of the given numbers by the first; then extract that root of the quotient whose index is one greater than the required number of means :

1. To insert six geometrical means between the numbers 3 and 384, we make m = 6, whence from the formula,

hence, we deduce the progression

3: 6 : 12 24 48 96: 192 : 384.

2. Insert four geometrical means between the numbers 2 and 486. The progression is

REMARK. When the same number of geometrical

inserted between each two of the terms of a

means are

geometrical progression, all the progressions thus formed will, when taken together, constitute a single progression.

Progressions having an infinite number of terms.

192. Let there be the decreasing progression

S=

arn

which expresses the sum of n terms, can be put under the form

Now, since the progression is decreasing, r is a proper frac tion, and TM is also a fraction, which diminishes as n increases. Therefore, the greater the number of terms we take, the more

will

1

α

× TM” diminish, and consequently, the nearer will the

sum of these terms approximate to an equality with the first n

part of S; that is, to

a

1 r

-

Finally, when ʼn is taken greater

will be less than any assignable number, or will become equal

to 0; and the expression

а

will represent the true value of

1

the sum of all the terms of the series. Hence,

The sum of the terms of a decreasing progression, in which the number of terms is infinite, is

This is, properly speaking, the limit to which the partial sums approach, as we take a greater number of terms of the progression. The number of terms may be taken so great as to

make the difference between the sum, and

1

а

as small as

we please, and the difference will only become zero when the number of terms taken is infinite.

What is the error, in each example for n =

Indeterminate Co-efficients.

= 4, n = 5, n = 6?

193. An IDENTICAL EQUATION is one which is satisfied for any values that may be assigned to one or more of the quantities which enter it. It differs materially from an ordinary equation.

The latter, when it contains but one unknown quantity, can only be satisfied for a limited number of values of that quan tity, whilst the former is satisfied for any value whatever of the indeterminate quantity which enters it.

It differs also from the indeterminate equation. Thus, if in the ordinary equation

ax+by+cz + d= 0

values be assigned to x and y at pleasure, and corresponding values of z be deduced from the equation, these values taken together will satisfy the equation, and an infinite number of sets of values may be found which will satisfy it (Art. 88). But if in the equation

ax+by+cz + d = 0,

we impose the condition that it shall be satisfied for any values of x, y and z, taken at pleasure, it is then called an identical equation.

194. A quantity is indeterminate when it admits of an infinite number of values.