the basket be placed three yards behind the first apple, that he could not gather them singly into it in an hour. What distance would he have to travel

(starting at the basket) to take them up?

Ans. 5 miles, 1700 yards.

3. If the first term of an arithmetical series be 1,. the last term 2, and number of terms 100; required, the common difference and sum of the series.

Ans. Sum 150, common difference.

4. The sum of the series is 1000, number of terms 1000, and last term 100; required, the first term, and common difference.

Ans. 98, common difference.

5. What is the difference between the sum of the first 1000 natural numbers, taken in order from unity, and the sums of the first 100, first and second hundreds, second and third hundreds, &c, to the ninth and tenth hundreds added together?

Ans. 450. 6. Required, the sum of 1000 terms of the series of odd numbers, 1, 3, 5, 7, 9, &c.

Ans. 1000000.

7. What would the first term of a decreasing series and its sum be, when the number of terms and common difference are 1000 and 3, and last term 1?

Ans. 2998 first term, and 1499500 sum.

8. Given, the sum of the series 8000, number of terms-4000, and common difference 2, to find the last term. Ans, 4001.

GEOMETRICAL PROGRESSION.

When a number of quantities increase by the same multiplier, or decrease by the same divisor, they form a Geometrical series.

This common multiplier, or divisor, is called the common ratio.

Thus, 1, 2, 4, 8, 16, &c.

Or, 1, TO, TOO, TOOO, &c.

Or, in general, a, ar, ar2, ar3, &c. is a geometrical series. And here r may be either a whole number, or a fraction. If r be greater than 1, the series will increase; but if it be less than 1, the series will decrease.

Let a denote the first term.

r the common multiplier, or ratio.

n the number of terms.

z the least term.

s the sum of all the terms.

Then the series will be a, ar, ar2, ar3, ar1 Za

....

And since it appears that the exponent of r, in any term, is one less than the number expressing the place of that term, it is evident, that zar"-1,

....

+arn-1.

Now, s=a+ar+ar2 + ar3 Therefore, multiplying both sides of the equation by r, we have rs-ar+ar2+ar3... arn-1+ar".

Let the first of these equations be subtracted from the second, and we have rs-s--a+ar", or (r-1)s =n{r"—1).

Therefore, s=

a(r_1)

r-1

Hence it appears, that the relations of the five quantities, a, r, n, z, s, to one another, are expressed by the two equations z=a(r”-1), s= a(r"—1)

r-1

And by these, having given any three of the four quantities, a, r, n, s. the remaining quantity may be found by the resolution of equations. If, however, n be not a small number, the cases of this problem will be most conveniently resolved by logarithms.

If we suppose the series to decrease, (in which case r will be a fraction,) and also that the number of terms is indefinitely great, then the formula for finding the sum, in this case, may be investigated as follows:

From the first equation we have rn= —,

a

and there

-1, and a(r”—1)=rz—a. There

fore, from the 2d equation, we find s=

ther, since r 1, s=2; but when the number of

1-r

terms is indefinitely great, and the series decreases, z is less than any assignable quantity, and is therefore to be reckoned as 0, or =0; therefore, in this.

EXAMPLE.

It is required to find the sum of the infinite series: 1, 2, 4, 1, &c. to infinity.

From the preceding doctrine we gather, that the constituent parts of a geometrical series are five, as well as in Arithmetical Progression; and that of the five, any three being given, the rest are had by an easy process, as will be shewn by formulas adapted to every case. We observed before, that of five things, ten combinations of every three of them may be made, which will all be different; hence also arises

twenty problems, which shew four different ways of finding each of the five parts. The formulas arise principally out of these two, viz. z=(-1) xa and

S=

a(r"—1), By what is called the resolution of

r-1

equations.

In order to illustrate numerically that the above formulas, or equations, are true, let us suppose the series 2, 6, 18, given,

Then the first term a will be

=2

26

By substituting in the first of the above equations, the numbers answering to the letters, then 182(33-1), and 26= 2(33-1) and so of all the rest,

3-1

NOTE. I have observed (as I before noticed, in Arithmetical Progression) that in several books, sets of formulas are to be found for each case; but I have not seen one perfect set. I confidently trust, that I have fully remedied the evil, by introducing correct and general formulas for each case.