357. From hence it appears that the sum of any geometrical progression may be found by multiplying the last term by the ratio of the progression, and dividing the difference between this product and the first term, by the difference between 1 and the ratio. In the example given in the preceding article, the last term is 512, the first term 1, and the ratio 2; and we have already found the sum to be 2 × 512-1 1024-l or 2-1 =1023. Now let there be a geometrical progression of 7 terms, of which the first term is 3 and the ratio 2, and let it be required to find the sum of the whole progression, we shall here have 358. Ifr be a proper fraction, then will r together with its powers be less than 1. In that case it will be more convenient for the purposes of calculation, to transform the above For instance, if it were required to find the sum of five terms of the decreasing progression 1, 4, 4, we shall have, a=1 n=5 and from Sa—ar” r = + 1-r we have S-1-3 5 31 = 16 359. Since we have already found that the last term, or lar, if we multiply this equation by r, we have rlar". From this last equation, any three of the four quantities S, a, r, l, being given, the fourth may be found. For -a r-1 S=rl ; a=rl-(r-1). S; r=' S-a (r-1). S+a and l r. EXAMPLES. EXAMPLES. Ex. 1. The sum of a geometrical progression, whose first term is 1, and last term 128, is 255. What is the common ratio? Ex. 2. Find the sum of a geometrical progression, whose first term is, common ratio, and number of terms 5. Ex. 3. If, in a geometrical progression, the sum of which is , the last term be, and the common ratio, what is the first term? Ex. 4. Find the sum of the series, an-1+ban-2 +b2an−3+ &c.....+b-1 to n terms. Ex. 5. There are three numbers in geometrical progression. The sum of the first and second is 15, and the sum of the first and last is 25. What are the numbers ? Let x the first, and y=the common ratio. Then the numbers are, x, xy, xy2. By the question, x+xy =15, and x+y=25. Taking the positive value, and resuming the equation x.(1+y)=15, we have 3x=15 and x=5. Therefore the numbers are, 5, 10, 20. Ex. 6. A servant agreed with his master to serve him for twelve months, upon this condition, that for his first month's service he should receive a farthing, for the second a penny, for the third four-pence, and so on; what did his wages amount to at the expiration of his service? ANS. £5,825. 8s. 54d. Ex. 7. There are three numbers in geometrical progression, whose sum is 52, and the sum of the first and second is to the sum of the first and third as 2: 5. Required the numbers? ANS. 4, 12, 36. Ex. 8. A bookseller being asked the value of his library, answered, I will sell it upon this condition, that I receive on the first of next month one shilling, on the first day of the following month two shillings, and so on for the space of twenty-four months. The bargain being struck, what did the library amount to? ANS. £479,260. 15s. CHAP. XLV. OF THE SUMMATION OF FRACTIONAL SERIES IN GEOMEAD INFINITUM, AND OF THE TRICAL PROGRESSION METHOD OF FINDING DECIMALS. THE VALUE OF CIRCULATING 360. WE have seen in the preceding Chapter, that when r is a proper fraction, the general expression for the sum of the progression is Now if this series be continued а - ар to any great number of terms, the value of " will become exceedingly small with respect to a, and if the series be continued ad infinitum, its value will be indefinitely small, and may therefore be rejected, and the approximating value of S may be expressed by a Let it be required, for example, to find the value of the series, 1+++++++ &c. ad infinitum. 361. The accuracy of this expression may be shewn by pursuing the same method, for finding the sum of a series ad infinitum, as we have in the preceding Chapter given for the summation of progressions, where the number of terms are limited. For instance, in the foregoing example, S=1+++++++ &c. ad infinitum. Multiplying by, we have, 1 S=1+++++ &c. ad infinitum. Subtracting this equation from the first, there remains S=1 and S2 for the sum required. b In general let a be the first term, and the ratio, where C C being a proper fraction, is less than one, and consequently c greater than b. The sum of the progression will be, ab ab2 abs S=a+ + + + &c. ad infinitum. C c2 C3 ab abs abs S= + + + &c. ad infinitum. b and therefore if the fractional ratio - be generally represented 362. In the same manner may be found the sums of progressions, the terms of which are alternately affected by the signs and. Let us take for example the series, b .S=a..S=. α b and representing - by r, the general expression will be S= C If therefore the series be, 3 6 12 24 α 1+r° |