2. Find the 100th term of the series

2.9.16.23.

3. Find the sum of 100 terms of the series

695.

Ans. 1.3.5.7.9...

Ans. 10000.

4. The greatest term is 70, the common difference 3, and the number of terms 21, what is the least term and the sum of the series? Ans. Least term 10: sum of series 840.

5. The first term of a decreasing arithmetical progression is 10,

1 3'

the common difference and the number of terms 21: required

the sum of the series.

Ans. 140.

6. In a progression by differences, having given the common difference 6, the last term 185, and the sum of the terms 2945, find the first term, and the number of terms.

Ans. First term 5, number of terms 31.

7. Find 9 arithmetical means between each antecedent and consequent of the progression 2.5.8.11.14...

Ans. Ratio, or r=0, 3.

8. Find the number of men contained in a triangular battalion, the first rank containing 1 man, the second 2, the third 3, and so on to the nth, which contains n. In other words, find the expression for the sum of the natural numbers 1, 2, 3 . . ., from 1 to n, inclu. n(n+1) sively.

Ans. S

2

9. Find the sum of the n first terms of the progression of uneven numbers 1, 3, 5, 7, 9 ... Ans. S=n2.

10. One hundred stones being placed on the ground, in a straight line, at the distance of 2 yards from each other: how far will a person travel, who shall bring them one by one to a basket, placed at 2 yards from the first stone?

Ans. 11 miles, 840 yards.

Geometrical Progression, or Progressions by Quotients.

221. A Geometrical progression, or progression by quotients, is a series of terms, each of which is equal to the product of that which precedes it, by a constant number, which number is called the ratio of the progression; thus in the two series :

each term of the first contains that which precedes it twice, or is equal to double that which precedes it; and each term of the second is contained in that which precedes it four times, or is a fourth of that which precedes it; they are then progressions by quotients, of

which the ratio is 2 for the first, and

1

4

for the second.

Let a, b, c, d, e, f... be numbers in a progression by quotients, they are written thus ; abc:def:g . . ., and it is enunciated in the same manner as a progression by differences; however it is necessary to make the distinction that one is a series of equal differences, and the other a series of equal quotients or ratios, in which cach term is at the same time an antecedent and a consequent, except the first, which is only an antecedent, and the last, which is only a consequent.

222. Let q denote the ratio of the progression

a:b:c:d...,

4 being >1 when the progression is increasing, and q<1 when it is decreasing: we deduce from the definition, the following equalities. b=aq, c=bq=aq2, d=cq=aq3, e=dq=aq...

and in general, any term n, that is, one which has n 1 terms be. fore it, is expressed by aq"-1.

Let be this term; we have the formula l=aq"-1, by means of which we can obtain the value of any term without being obliged to find the values of all those which precede it. That is, the last term of a geometrical progression is equal to the first term multiplied ́

by the ratio raised to a power whose exponent is one less than the

number of terms.

For example, the 8th term of the progression

is equal to 2×37=2×2187=4374.

2:6:18:54..

In like manner, the 12th term of the progression

223. We will now proceed to determine the sum of n terms of

the progression

nth term.

k:l, l

a:b:c:def:...:i:k: l, 7 denoting the

We have the equations (Art. 222),

b=aq, c=bq, d=cq, e=dq,... k=iq, l=kq;

and by adding them all together, member to member, we deduce

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

or, representing the required sum by S,

...

+i+k)q;

S-a (S-1)q=Sq-lq, or Sq-S=lq-a;

That is, to obtain the sum of a certain number of terms of a progression by quotients, multiply the last term by the ratio, subtract the first term from this product, and divide the remainder by the ratio diminished by unity.

When the progression is decreasing, we have q<1 and l<a; a-lg the above formula is then written under the form S=1-q

in order that the two terms of the fraction may be positive. By substituting aq"-1 for 7 in the two expressions for S, they be

EXAMPLES.

1. Find the first eight terms of the progression

2:6:18: 54: 162 : 2×37 or 4374

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.

3. What debt may be discharged in a year, or twelve months, by paying $1 the first month, $2 the second month, $4 the third month, and so on, each succeeding payment being double the last; and what will be the last payment?

Ans. Debt, $4095: last payment, $2048.

This result, which is sometimes a symbol of indetermination, is also often a consequence of the existence of a common factor (Art. 113), which becomes nothing by making a particular hypothesis respecting the given question. This, in fact, is the case in the present question; for the expression q′′-1 is divisible by q—1, (Art. 59), and gives the quotient

We can obtain the same result by going back to the proposed progression, a : b:c:...:l, which, in the particular case of q=1, reduces to a: a: a: . : a, the sum of which series is equal

to na.

0

The result given by the formula, may be regarded as indicating that the series is characterized by some particular property. In fact, the progression, being entirely composed of equal terms, is no more a progression by quotients than it is a progression by diffeTherefore, in seeking for the sum of a certain number of

rences.

the terms, there is no reason for using the formula S=

a(q"—1)

9-1

in preference to the formula S=

(a+1)n

2

which gives the sum in

the progression by differences.

Of Progressions having an infinite number of terms.

225. Let there be the decreasing progression a:b:c:d e:f:...,

a-aq" 1-q

containing an indefinite number of terms. The formula S=

which represents the sum of n of its terms, can be put under the

Now, since the progression is decreasing, q is a fraction; and qr is also a fraction, which diminishes as n increases.

Therefore the

greater the number of terms we take, the more will

a

diminish, and consequently, the more will the partial sum of these terms approximate to an equality with the first part of S, that is, to