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Examples of Substitution of Unknown Quantities.

or y

Dividing, then, by yć, we have

12 + 13 -4q=0; which is satisfied by the value of q,

g=0; or dividing by 9, we have

12 4 + 13 –4= 0, whence

q=+), orq=ŁV-4. Now the first of the given equations becomes, by the substitution of


qoys + q3 y5=5; hence, by the substitution of the above values of q, we have

y 60, 1 = = 0 x 20 8 indeterminate;

+ 2,x=1;
or y=+v Š XV - 3, x=V 20.

3. Solve the two equations

31 x4 + 9x2 y2 = 20 yé, (y2-y)2+(3x y+2y)2—9 x2(2y+3)—12 y(x+2y)=0. Ans. I= = 0,

and y=0;

- 11},
- 315,

: ,


y -1; or r=1(-5V-5,) and y=1EV-5;

or x=#V-5, and y=1. 4. Solve the two equations

23 + 2 x y = 3,
x y2 + 2xy = 3.

Ans. x=1, and y=1.



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Examples of Substitution of Unknown Quantities.

5. What two numbers are they, twice the sum of whose squares is 5 times their product, and the sum of whose sixth

Ans. 2 and 1, or -2 and 1.

powers is 65.

6. What two numbers are they, the difference of whose fourth powers is 65, and the square of the sum of whose

Ans. + 2, and £3.

squares is 169.


To find the last Term.




Arithmetical Progression.

244. An Arithmetical Progression, or a progression by differences, is a series of terms or quantities which continually increase or decrease by a constant quantity.

This constant increment or decrement is called the common difference of the progression.

Throughout this section the following notation will be retained. We shall use

a = the first term of the progression,
1= the last term,
r= the common difference,
n = the number of terms,
S= the sum of all the terms.

245. Problem. To find the last term of an arithmetical progression when its first term, common difference, and number of terms are known.

Solution. In this case a, r, and n, are supposed to be known, and l is to be found. Now the successive terms of the series if it is increasing are

a, a tr, a +2r, at 3r, a +4r, &c.;

Sum of two Terms equally distant from the extremes.

so that the nth term is obviously

1= a + (n − 1)r. But if the series is decreasing, the last term must be

1=a-(n-1) r. Both these cases are, however, included in one, if we suppose r to be negative when the series is decreasing.

246. Corollary. In like manner any other term, such as the mth, is

a + m - 1).

247. Corollary. By writing the series in an inverted order, beginning with the last term, a new series is found, of which the first term is l, and the common difference Hence the mth term of this series, that is, the mth term counting from the last of the given series, is

1- (m — 1)r.


248. Corollary. The sum of the mth term and of the mth term from the last is, therefore,

[a + (m — 1)r] + [? — (m — 1)r] =a+l; that is, the sum of any two terms, taken at equal distances from the two extremes of an arithmetical series, is equal to the sum of the two extremes.

249. Problem. To find the sum of an arithmetical progression when its first term, last term, and number of terms are known.

Solution. In this case, a, l, and n are supposed to be known, and S is to be found.

To find the Sum of the Progression.


Suppose the terms of the series to be written as follows, first in the regular order, and then in an inverted order : a, b, c,

i, k, l; 1, k, i, .

c, b, a. The sum of the terms of each of these progressions being S, the sum of both of them must be 2 S, that is,

28=(a+1)+(6+k)+(c+i)...titc)+(2+6)+(H+a). But by the preceding corollary, we have

atl=6+k=cti=&c. Hence 2 S is equal to as many times (a + ) as there are terins in the series, that is,

28= (a +1);

or S=i (at)n; that is, the sum of a progression is equal to half the sum of the two extremes, multiplied by the number of terms. 250. Corollary. From the equations

1= a + (n − 1)r,

S=1 (a +i)n; either two of the quantities a, l, r, n, and S can be determined when the other three are known.


1. Find the 100th term of the series 2, 9, 16, &c.

Ans, 695. 2. Find the sum of the preceding series.

Ans. 34850.

3. Find S, when a, r,

and n are known. Ans. S=1 [2 a +(n-1)r] n.

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