B starts a days later, and travels b miles per day. How long will A travel before he is overtaken by B?

For instance, in the preceding example, if B had started one day later, he could never have overtaken A.

Ex. 19. A traveler set out from a certain place and went 1 mile the first day, 3 the second, 5 the third, and so on. After he had been gone three days, a second traveler sets out, and goes 12 miles the first day, 13 the second, and so on. In how many days will the second overtake the first?

Ans. In 2 or 9 days. Let the student illustrate this example by a diagram like the preceding.

GEOMETRICAL PROGRESSION.

(240.) 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.

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.

(241.) To find the last term of a geometrical progression. Let a represent the first term of the progression, and r the common ratio; then the successive terms of the series will be a, ar, ar2, ar3, ar1, &c.

The exponent of r in the second term is 1, in the third term

is 2, in the fourth term 3, and so on; hence the nth term of the series will be

ar"-1.

If, therefore, we put 7 for the last term, and n the number of terms of the series, we shall have

That is,

l=ar-1.

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.

(242.) To find the sum of all the terms of a geometrical progression.

If we take any geometrical series, and multiply each of its terms by the ratio, a new series will be formed, of which every term except the last will have its corresponding term in the first series. Thus, take the series

1, 2, 4, 8, 16, 32,

the sum of which we will represent by S, so that

S=1+2+4+8+16+32.

Multiplying each term by 2, we obtain

2S 2+4+8+16+32+64.

The terms of the two series are identical, except the first term of the first series and the last term of the second series. If, then, we subtract one of these equations from the other, ali the remaining terms will disappear, and we shall have

2S-S=64-1.

In order to generalize this method, let a, ar, ar1, &c., represent any geometrical series, and S its sum; then

S=a+ar+ar2+ar+......+are+ar"-.

Multiplying this equation by r, we have

rS=ar+ar2+ar3+ar*+......+ar"-1+ar".

Subtracting the first equation from the second, we obtain

or, substituting the value of I already found, we shall have

Hence, to find the sum of the terms of a geometrical progression,

Multiply the last term by the ratio, subtract the first term, and divide the remainder by the ratio less one.

If the series is a decreasing one, and r consequently represents a fraction, it is convenient to change the signs of both numerator and denominator in this expression, which then be

of which any three being given, the other two may be found. Accordingly, as in arithmetical progression, 20 different cases may arise, all of which are readily solved, with the exception of those in which n is the quantity sought. The value of n can only be found by the solution of an exponential equation. See Art. 352. These different cases are all exhibited in the following table for convenient reference.

Ex. 3. Given the first term 2, the ratio 3, and the number of terms 10, to find the last term.

Ans. 39366.

Ex. 4. Given the first term 1, the last term 512, and the sum of the terms 1023, to find the ratio.

Ex. 5. Given the last term 2048, the number of terms 12, and the ratio 2, to find the first term.

Ex. 6. A person being asked to dispose of his horse, said he would sell him on condition of receiving one cent for the first nail in his shoes, two cents for the second, and so on, doubling the price of every nail to 32, the number of nails in his four shoes. What would the horse cost at that rate?

Ans. $42,949,672.95

(244.) To find any number of geometrical means between two given numbers.

In order to solve this problem, it is necessary to know the ratio. If m represent the number of means, m+2 will be the whole number of terms. Substituting m+2 for n in Formula 9 Art. 243, we obtain

m+1

r= V a

That is, to find the ratio, divide the last term by the first term, and extract the root denoted by the number of means plus one. When the ratio is known, the required means are obtained by continued multiplication.

Ex. 1. Find three geometrical means between 2 and 162. Ex. 2. Find two geometrical means between 4 and 256.

(245.) Of decreasing progressions having an infinite number of terms.

In a decreasing progression, since r is a proper fraction, r is less than unity, and the larger the number n, the smaller will be the quantity". If, therefore, we take a very large number of terms of the series, the quantity r", and, consequently

will be very small; and if we take n greater