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91. If 4 quantities are proportionals, the like powers of these quantities are also proportionals.

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Multiplying equals by equals

X &c. m times = &c.

с

d

с
с
X X
d

d

(Art. 49.)

α

b

m times.

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92. If there be any number of equal ratios, then will one antecedent be to its consequent, as the sum of all the antecedents to the sum of all the consequents.

Let abcdef

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then will ab::a+c+e: b+d+f

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.. ab+ad+af ba+be+be (Art. 16.)
Or, a(b+d+f)=b(a+c+e)
..aba+c+e: b+d+f (Art. 84.)

The same proof will extend to any number of

ratios.

ARITHMETICAL PROGRESSION.

93. Def.-Quantities are in Arithmetical Progression when they increase or decrease by a common difference.

As 1, 4, 7, 10, 13, &c.

or

a, a+d, a +2d, a+3d, &c. or a, a-d, a-2d, a-3d, &c.

Hence, if a represent the first term, d the common difference, the progression is a, a+d, a+2d, a+3d, &c., in which the coefficient of d is in each term one less than the number of terms in the progression; therefore, if there are n terms, the last term l=a+(n−1)d, and of the 4 quantities a, l, n, d, we may determine any one of them, from the other 3 being given.

Ex. The first term of an Arithmetical Progression is 1, the last 5, and the number of terms 9; find the common difference.

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.. the progression is 1, 1, 2, 2, 3, &c.

94. The sum of an Arithmetical Progression is found by multiplying the sum of the first and last terms, by half the number of terms.

Let a +(a+d)+(a+2d) &c. . . +l=S
l+(l−d)+(l−2d) &c. . . +a=S
a+1+(a+1)+(a+1) &c. to n terms =2S
(a+1)n=2S

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Ex. Find the sum of 100 terms of the pro

gression 1, 2, 3, 4, 5, &c.

95.

Here a=1, l=100, n=100

n

..S=(a+1)2 =(1+100)50=5050

Since la+ (n-1)d; (Art. 93.)

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By this last equation, any 3 of the terms S, a, d, n, being given, the remaining one can be determined.

Ex. 1. Find the sum of 100 terms of the progression 3, 5, 7, 9, 11, &c.

Here a=3, d=2, n=100

.. S=[2a+(n− 1)d] 1⁄2=10200

Ex. 2. Find the sum of 20 terms of the progression 11, 8, 5, 2, −1, −4, &c.

Here a=11, d=-3, n=20

... S=[2a+(n − 1)d]

n

== - 350

2

Ex. 3. The sum of a certain number of terms of an Arithmetical Progression is 240, the common difference 2, and the first term 2; required the number of terms.

Here S=240, d=2, a=2

by substituting these numbers

n2+n=240

By solving this quadratic equation

n=15 the number required.

96. Find 3 Arithmetical Means or intermediate terms of an Arithmetical Progession between 214 and 282. The progression is 214, 214+d, 214+2d, 214+3d, 214+4d, and the last term is 282.. 214+4d=282 .. d=17 the common difference. The intermediate terms are therefore 231, 248, 265.

GEOMETRICAL PROGRESSION.

97. Quantities are in Geometrical Progression, when each term of the progression is a certain multiple or part of the preceding one. The con

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