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The first of these is the coefficient of the second term; the coefficient of the second multiplied by g forms the coefficient of the third term, &c.

7 × 10 = 21. 21 X // = 35.

Now 35 multiplied by = 1 will not be altered; hence two successive coefficients will be alike. 21 multiplied by produced 35; so 35 multiplied by must reproduce 21. In this way all the terms will be reproduced; for the last half of the fractions are the first half inverted.

This demonstration might be made more general, but it is not necessary.

XLVI. Progression by Difference, or Arithmetical Progression.

A series of numbers increasing or decreasing by a constant difference, is called a progression by difference, and sometimes an arithmetical progression.

The first of the two following series is an example of an increasing, and the second of a decreasing, progression by dif

ference.

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It is easy to find any term in the series without calculating the intermediate terms, if we know the first term, the common difference, and the number of that term in the series reckoned from the first.

Let a be the first term, r the common difference, and n the number of terms. The series is

a, a +r, a + 2 r, a + 3r....a + (n − 2) r, a + (n − 1) r.

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The points.. are used to show that some terms are left out of the expression, as it is impossible to express the whole until a particular value is given to n.

Let be the term required, then

l= a + (n − 1) r.

Hence, any term may be found by adding the product of the common difference by the number of terms less one, to the first

term.

Example.

What is the 10th term of the series 3, 5, 7, 9, &c.

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What is the 13th term of the series 48, 45, 42, &c.?

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12.

748+ (12 X-3)= 48-36

Let a, b, c, be any three successive terms in a progression by difference.

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That is, if three successive terms in a progression by difference be taken, the sum of the extremes is equal to twice the

mean.

Example.

Let the three terms be 3, 5, and 7.

2 × 5=7+3=10.

Example 2d. Let 7 and 17 be the first and last term, what is the mean?

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Let a, b, c d, be four successive terms of a progression by difference.

b-a-d-c

b+c=a+d.

That is, the sum of the two extremes is equal to the sum of the two means.

Example.

Let 5, 9, 13, 17, be four successive terms.

9+1317+5=22.

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Let a, b, c, d, e h, i, k, l, be any number of terms in a progression by differences; by the definition we have b―ac-bd-c-e-d-i-hk-i-l—k.

=

b-a l-k

c-b ki
=

d-cih, &c.

which by transposition give

a+l=b+k,

b + k = c + i.

c+ i=d+h, &c.

=

That is, if the first and last be added together, the second and the last but one, the third and the last but two, the sums will all be equal.

Example.

Let 3, 5, 7, 9, 11, 13, be such a series.

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It will now be easy to find the sum of all the terms in any progression by difference, and that even when but part of the terms are known.

Let S represent the sum of the series, then we have
S=a+b+c+d+....h + i + k + l.

Also S1+k+i+h+....d+c+b+a.
Adding these term to term as they stand,

28=(a+1)+(b+ k) + (c + i) + (d+h) + ........ (d + h)+(c+ i) + (b+k)+(a+1)

But it has just been shown that

a+l=b+k = c + i, &c.

That is, all the terms are now equal, and one of them being multiplied by the whole number of terms will give the whole sum thus

28=n (a + 1)

S = n (n + 1)

2

Hence, the sum of a series of numbers in progression by difference is one half of the product of the number of terms by the sum of the first and last terms.

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Example.

strokes does the hammer of a clock strike in 12

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In the formula 1 = a + (n r to represent the difference; thus

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contain five different things, viz. a, l, n, d, and S, any three of which being given, the other two may be found, by combining the two equations. I shall leave the learner to trace these himself as occasion may require.

Examples in Progression by Difference.

1. How many strokes do the clocks of Venice, which go on to 24 o'clock, strike in a day?

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, 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, (24990 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?

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