2. Suppose 100 stones to be placed in a straight line 3 yards asunder; how far would a person travel who should set a basket 3 yards from the first, and then go and pick them up one by one, and put them into the basket?

3. After A, who travelled at the rate of 4 miles an hour, had been set out 2 hours, and B set out to overtake him, and in order thereto went four miles and a half the first hour, four and three fourths the second, five the third, and so on, increasing his rate one fourth of a mile each hour. In how many hours will he overtake A ?

The above example is solved by using both the above formulas. The known quantities are the first term, the difference, and the sum of all the terms. The unknown are the last term, and the number of terms. It involves an equation of the second degree. It is most convenient to use x, y, &c. for the unknown quantities.

4. A and B set out from London to go round the world, (23661 miles,) one going East and the other West. A goes one mile the first day, two the second, three the third, and so on, increasing his rate one mile per day. B goes 20 miles a day. In how many days will they meet, and how many miles will each travel?

5. A traveller sets out for a certain place, and travels 1 mile the first day, 2 the second, and so on. In 5

days afterwards another sets out, and travels 12 miles a day. How long and how far must he travel to overtake the first?

6. A and B 165 miles distant from each other set out with a design to meet; A travels 1 mile the first day, 2 the second, 3 the third, and so on. B travels 20 miles the first day, 18 the second, 16 the third, and so on. How soon will they meet?

Ans. They will be together on the 10th day, and continuing that rate of travelling, they may be together again on the 33d day. Let the learner explain how this can take place.

7. A person makes a mixture of 51 gallons, consisting of brandy, rum, and water; the quantities of which are in arithmetical progression. The number of gallons of brandy and rum together, is to the number of gallons of rum and water together as 8 to 9. Required the quantities of each.

Let x the number of gallons of rum

and y = the common difference.

Then xy, x, and x+y will express the three quantities.

8. A number consisting of three digits which are in arithmetical progression, being divided by the sum of its digits, gives a quotient 48; and if 198 be subtracted from the number, the digits will be inverted. Requir ed the number.

9. A person employed 3 workmen, whose daily wages were in arithmetical progression. The number of days they worked was equal to the number of shillings that the second received per day. The whole amount of their wages was 7 guineas, and the best workman received 28 shillings more than the worst. What were their daily wages?

Progression by difference is only a particular case of the series by difference, explained Arts. XL. and XLI. All the principles and rules of it may be derived from the formulas obtained there. It would be a good exercise for the learner to deduce these rules from those formulas.

XLVII. Progression by Quotient, or Geometrical Progression.

Progression by quotient is a series of numbers such, that if any term be divided by the one which precedes it, the quotient is the same in whatever part the two terms be taken. If the series is increasing, the quotient will be greater than unity, if decreasing, the quotient will be less than unity.

The following series are examples of this kind of progression.

In the first the quotient (or ratio, as it is generally called,) is 2, in the second it is .

Let a, b, c, d, .... k, l, be a series of this kind, and let q represent the quotient.

Putting successively the value of b into that of c, and

that of c into that of d &c., they become

b = a q, c = a q3, d = a q3, c = a q1

designating by n, the rank of the term 7, or the number

of terms in the proposed progression.

Any term whatever in the series may be found without finding the intermediate terms, by the formula

1 = a qn−1.

Example.

What is the 7th term of the series 3, 6, 12, &c ?

Here a = 3, q = 2, and n

- 1 = 6.

Ans. 192.

1 = 3 × 26 = 192.

We may also find the sum of any number of terms

of the progression

a, b, c, d, &c.

If we add the equations

b = a q, c = bq, d = c q, e = d q . . . . . l = q,

we obtain

b+c+d+e+....l = (a+b+c+d+e+ .... k) q.

Observe that the first member is the sum of all the terms of the progression except the first, a, and the part of the second member enclosed in the parenthesis, is the sum of all the terms except the last, ; and this, multiplied by q, is equal to the first member.

Now putting S for the sum of all the terms, we have

b+c+d+e+..... l = S-a α a+b+c+d+e+.....k=S-l.

Hence we conclude that