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This is evident, since the last term consists of the sum of the first term, and a certain number of times the common difference; l-a, expresses that certain number of times the common difference, and this remainder being divided by the number the common difference is multiplied by, the quotient is necessarily the common difference.

Again, if in a decreasing progression we have la-(n—1) d, adding to each side of the sign of equality (n-i) d, then (n-1) d +1=a, from which I being subtracted from these equal sums, we have (n-1) da-l, and dividing these equal remainders by a-l (n-1), it is found that d=

n-l

Here the common difference is equal to the first term minus the last, divided by the number of terms minus 1.

Example 1. Suppose an A.P. of 14 terms, in which the first term a=3, and the last 155, it is required to find the common difference d.

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and the A.P. is 3, 7, 11, 15, 19.........55.

Ex. 2. What is the common difference of a series consisting of 10 terms, the first term is 11 and the last 4?

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and the equi-difference is 11, 103, 43, 88, 73.......4.

358. In a similar manner do we determine n, the number of terms, by means of a, the first term, 7 the last, and d the common difference.

As before, la+(n-1) d.

Let a be subtracted from each side of the sign of equality, then la(n-1) d, dividing both equal remainders by d, it follows that =n-1, and adding 1 to both equal quantities, we

have 1+

l-a
d

-a

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Hence the number of terms is found by dividing the last term, minus the first, by the common difference and adding 1 to the quotient. For taking away the first term from the last, the remainder is the common difference multiplied by a certain number, which being divided by the common difference, the quotient is the

number by which the common difference was multiplied; this number is 1 less than the number of terms, therefore 1 must be added to the quotient.

For a decreasing A.P. we have: l-a-(n-1) d, adding to both sides of the sign of equality (n-1) d.

.. l+(n−1) d=a, substracting from both sides of the sign of equality, we have (n-1) d=a―l, which being divided by d, gives N -1 and adding to both sides 1,

d

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Hence the number of terms is found, &c.

359. If we were to subtract (n-1) d from both sides of l=a +(n-1) d, we have l-(n-1) d=a.

Therefore the first term of an A.P.=the last term, minus the product, of the common difference and the number of terms less 1.

This truth is rendered evident when we consider that the last term consists of two parts, the first term and the product of the common difference and the number of terms minus 1; then subtracting the second part, the remainder is the first term. When the A.P. is decreasing, we have:

l—a—(n—1) d, and adding (n—1) d to each side, l+(n−1) d = a. Hence the first term is found, &c.

Example 1. The last term is 53, the common difference 3, and the number of terms 18 Required the first term.

a=53—(18—1)3=2.

Ex. 2. The last term of an A.P. is 5, the common difference 2, and the number of terms 52. What is the first term?

a=5+(52-1)2=107.

360. An important case in A.P. is to find the sum of the terms; the operation of adding all the successive terms together would be very tedious; we shall now determine a method by means of which the sum of the terms of any A.P. can at once be found.

Let it be required to ascertain the sum of the terms of the following A.P.

3, 7, 11, 15, 19, 23, 27, 31, 35, 39.

In order to determine the sum of the terms, let us write the

progression, both as an increasing and decreasing series, setting the terms in order, and adding together the corresponding ones, as follows:

Increasing series... 3, 7, 11, 15, 19, 23, 27, 31, 35, 39. Decreasing series...39, 35, 31, 27, 23, 19, 15, 11, 7, 3. 42, 42, 42, 42, 42, 42, 42, 42, 42, 42.

We obtain every term 42, as the sum of two terms, and this sum is repeated as many times as there are terms in the progression; therefore 10 x 42=420 is the sum of both series, or twice the sum of the terms of the proposed progression. Whence the sum 10 × 42 required is =210. 2

a,

Let us proceed in a similar manner with any A.P. :

INCREASING SERIES.

:

a+d, a+2d, a+3d, a+4d, a+5d, a+6d, a+7d.

DECREASING SERIES.

a+7d, a+6d, a+5d, a+4d, a+3d, a+2d, a+d, a. 2a+7d, 2a+7d,2a+7d, 2a+7d, 2a+7d, 2a+7d,2a+7d,2a+7d.

In the first series every following term increases by d, whilst in the second it diminishes by d, The sum of the terms of both series is 8x (2a+7d), or the number of terms in one progression multiplied by the sum of any two corresponding terms; therefore 8x (2a+7d) the sum of the proposed series:

2

Were we to take the number of terms to be n, we should have:

a,

a+d,

INCREASING SERIES.

a+2d, a+3d...a+(n-2)d, a+(n-1)d.

DECREASING SERIES.

a+(n-1)d, a+(n-2)d, a+(n-3)d, a+(n-4)d,......a+d. a. 2a+(n-1)d, 2a+(n-1)d, &c.. Qa+(n−1)d.

The sum of every two corresponding terms is 2a+(n−1)d, which quantity being repeated n times, we have, supposing s to (2a+ (n-1)d) represent the sum of the terms of one series, 2s=n⋅ for the sum of both series; therefore the sum of the terms of any From which is derived the

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series is sn

(2a+(n-1)d)

2

following rule:

2

The sum of the terms of any A.P. is found by multiplying half the sum of the first and last terms by the number of terms in the series.

361. In this progression: 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, we may observe that the sum of the first and last term is 42; the sum of the second and last but one is likewise 42; of the third and eighth; of the fourth and seventh, and so on of any two terms equally distant from the first and last.

Also for any arithmetical series ;

a, a+d, a +2d, a+3d, a+4d......a+(n—4)d, a+(n-3)d, a+(n-2)d, a+(n−1)d.

The sum of the first and last term is equal to the sum of the second and last but one, &c., viz., 2a+(n−1)d.

The reason of this is manifest from what has been said, for, we find the sum of two terms, as just mentioned, to be the same as adding together the two corresponding terms of an increasing and decreasing series.

Therefore, if the sum of the first and last terms of a series,

n

a+l, be multiplied by half the number of terms, we obtain

2'

the sum of the terms of the series ; hence s=(a+1) {2a+(n−1)d} n

2

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Ex. 1. In an arithmetical series, the first term a=4, the common difference d=5, and the number of terms n=36, find the sum s.

8=

{2×4+(36-1)5}36_(8+175)36

2

2

=263 x 18=4734.

Ex. 2. Given the first term a=3, the last term 1=36, and the number of terms n=18, to find the sum of the A.P.:

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362. In the following propositions will be found the different cases which present themselves in A.P., they are deductions from what has been explained:

1. Given a, d, n, to find l. Ans. l=a+(n—1d.

d

2. Given a, d, s, to find l. Ans. 1=~;+

2ds +

2

2 s

3. Given a, n, s, to find l. Ans. 1=2 -a.

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