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Examples involving Arithmetical Progression.

4. Find four numbers in arithmetical progression whose sum is 28, and continued product 585.

Ans. 1, 5, 9, 13.

5. The sum of the squares of the first and last of four numbers in arithmetical progression is 200, and the sum of the squares of the second and third is 136; find them.

Ans. 2, 6, 10, 14.

6. Eighteen numbers in arithmetical progression are such, that the sum of the two mean terms is 311, and the product of the extreme terms is 85. Find the first term and the common difference.

Ans. The first term is 3,

the common difference is 11.

SECTION II.

Geometrical Progression.

257. A Geometrical Progression, or a progression by quotients, is a series of terms which increase or decrease by a constant ratio.

a, l, n, and S will be used in this section as in the last, to denote respectively the first term, the last term, the number of terms, and the sum of the terms; and r will be used to denote the constant ratio.

258. Problem. To find the last term of a geometrical progression when its first term, ratio, and number of terms are known.

To find the last Term and Sum.

Solution. In this case a, r, and n are given, to find l Now the terms of the series are as follows:

a, ar, ar2, ar3... &c. ... arn-1,

so that, the last or nth term is

1=ar"-1;

that is, the last term is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms.

259. Problem.

To find the sum of a geometrical progression, of which the first term, the ratio, and the number of terms are known.

Solution. We have

S=a+ar+ar2 + &c....+arn-2+arn−1.

If we multiply all the terms of this equation by r, we have rs=ar+are+ar3+ &c....+arn-1+arn; from which, subtracting the former equation, and striking out the terms which cancel, we have

or

whence

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(r− 1) S = ar" — a = a (r” — 1);

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Hence, to find the sum, multiply the first term by the difference between unity and that power of the ratio whose exponent is equal to the number of terms, and divide the product by the difference between unity and the ratio.

Examples in Geometrical Progression.

260. Corollary.

The two equations

l = a rn-1

(r — 1) S = a (r” — 1)

give the means of determining either two of the quantities a, l, r, n, and S, when the other three are known.

But it must be observed, that, since n is an exponent, it can only be determined by the solution of an exponential equation.

261. EXAMPLES

1. Find the 8th term and the sum of the first 8 terms of the progression 2, 6, 18, &c., of which the ratio is 3.

Ans. The 8th term is 4374,

the sum is 6560.

2. Find the 12th term and the sum of the first 12 terms of the series 64, 16, 4, 1, †, &c., of which the ratio is .

Ans. The 12th term is 536,

the sum is 85,65535

196608

3. Find S, when a, l, and r are known.

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4. Find the sum of the geometrical progression of which the first term is 7, the ratio, and the last term 12.

Ans. 121.

5. Find r and S, when a, l, and n are known.

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Examples in Geometrical Progression.

6. Find the ratio and sum of the series of which the first term is 160, the last term 38880, and the number of terms 6.

Ans. The ratio is 3,

the sum is 58240.

7. Find r, when a, 1, and S are known.

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8. Find the ratio of the series of which the first term is 1620, the last term 20, and the sum 2420.

Ans..

9. Find a and S, when 1, r, and n are known.

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10. Find the first term and sum of the series of which the last term is 1, the ratio, and the number of terms 5.

Ans. The first term is 16,

the sum is 31.

11. Find 7, when a, r, and S are known.

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12. Find the last term of the series of which the first

term is 5, the ratio, and the sum 6.

Ans. 21·

13. Find a, when 1, r, and S are known.

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14. Find the first term of the series of which the last

term is, the ratio, and the sum 6

Ans. 5.

Infinite Geometrical Progression.

15. Find a and 7, when r, n, and S are known.

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16. Find the first and last terms of the series of which the ratio is 2, the number of terms 12, and the sum 4095. Ans. The first term is 1,

the last term 2048.

262. An infinite decreasing geometrical progression is one in which the ratio is less than unity, and the number of terms infinite.

263. Problem. To find the last term and the sum of the terms of an infinite decreasing geometrical progression, of which the first term and the ratio are known.

Solution.

Sincer is less than unity, we may denote it by a fraction, of which the numerator is 1, and the denominator r' is greater than unity; and we have

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Since, then, the number of terms is infinite, the formulas

for the last term and the sum become

1 = a⋅rn-1 = a X 0

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=

ar!

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