The two other cases of formula (1) are when Irn or arn are given. The next two examples are on formula (2). Ex. 3. The common ratio is 4, and the sum of 5 terms is 682; find the number of terms. Ex. 5. If a population of 50,000 be increased in 2 years to 53,045, find what the population will be at the end of 5 years. Let r = the ratio which the population at the beginning of any one year bears to that at the beginning of the preceding year, Ex. 6. If Sn denote the sum of n terms of a G. P., whose common ratio is 2, and if San Sn = 129 find n. j' 1 (+1), jn = 128; 2n 128. = = 7. Examples LXVI. 1. Find the 20th term of the series 2, 4, 8, &c.; the 16th term of 5, 20, 80, &c.; the nth term of 3x, 9x2, 27x3, &c.; the nth term of &c.; the nth term of (a — x), (a — x)3, (a − x)3, &c.; the nth a 1 1 2. Sum the following Geometric series, terms; 5+ 10+20 +, &c., to 10 terms; 7+ 21 + 63 +, &c., to 19 +3 +2, &c., to 8 terms; 4+2 + 1, &c., to 16 terms and to n terms; 6 + 4 + 23, &c., to 30 terms and to 3n terms; + (a+ а3х +, &c., (a2 + ax + x2) + (a3 − x3) · ax3 + x1), &c., to (n + 1) terms; (a+x)2 + (a2 + (a − x)2 +, &c., to 2n terms; terms; (√√2 − 1) + 1 + (√2 + 1), &c., to n terms. 3. Insert 5 Geometric means between 2 and 128; also 8 G. M. between 4. If ten terms be in G. P., and the last be 1024 times the first, find the common ratio. Also find the relation between the 2nd and the 8th term. 5. If the 20th term of a series be 64, and the 24th term be find the 1st term and the 100th term. 625 4 6. The mth term of a Geometric series is a, and the nth term is b; find the first term. 7. The sum of the first three terms of a Geometric series is double of the first term: find the common ratio. 8. If the sum of x terms of a Geometric series be 4 times the sum of x terms, find the value of r. Also if the sum of five terms of a 2 Geometric series be 33 times the sum of five terms, find the value of r. Is the first term of the series determined in either of these cases? 9. The population of Great Britain in 1841 was 27,019,548; in 1851 it was 27,770,091. The ratio of increase being supposed uniform, find the increase each year in the population; also the rate of increase per cent. per annum. 10. If a sum of money be left to accumulate at compound interest, the principals of the successive years form a Geometric series. State what is the first term and what the ratio of each principal to the preceding one; also if 16,000l. be put to interest for 4 years, and amount to 19,4481. 28. at compound interest; find the rate per cont. per annum of interest. 11. If the areas of a series of squares be in G. P., prove that their sides will also be in G. P. 12. If Sn denote the sum of a series of n terms of a G. P., show that Sn + rsn-2 = (1 + r)s, -2 ; also show that S2+1 S = rsn' - 13. Find the sum of the following series to n terms; 1.2 +3.6 +9.18+, &c.; 1 +, &c. n 14. Show that the sum of the terms of a G. P. is less than (a + b),5, also that the product of n terms of an A. P. is greater than (ab) 2, a, denoting the 1st and nth terms respectively. b, 15. Find a number such that the Arithmetic mean between it and 20 shall be three times the Geometric mean. INFINITE GEOMETRIC SERIES. 155. If the value of r in a Geometric series be greater than unity, the terms continually increase, and if the number of terms be great enough, the last term may be made greater than any assigned number; and of course the sum of n terms will also be greater than any assigned number. It is therefore said that when r > 1 and n increases without limit, the nth term and the sum of n terms increase without limit. Ex. Take the series 3, 6, 12, &c. If a number, say ten million, be proposed, it is possible to take so many terms, that the last of them shall be greater than ten million, and similarly if any greater number be proposed. If r be equal to unity, it is evident that all the terms are alike, and the last term is not greater or less than any other term; but in this case the sum of n terms can be made as great as we please, by giving a sufficiently large value to n. If r be less than unity, the terms continually decrease, and if the number of terms be sufficiently great, the last or nth term may be made less than any assigned number. This may be seen from the general expression from the nth term, which is ar- 1 ̧ And it is evident that (1 + k)"−1 can be made as large as we please by taking n large enough, and therefore than any assigned number. a can be made less (1+ k)". Ex. Take the series 3, 2, 3, &c.; and let a small fraction, say 000000001, be proposed, it is possible to find a term of the series which shall be less than this. To find the sum of n terms in this case we have the formula We now proceed to discuss the value of s when n is made indefinitely Now the latter part of this expression (1) n becomes inde finitely small when n is made indefinitely great, or the difference between the true value of 8 and the quantity a may be made less 1 - 2 than any assignable number, by sufficiently increasing n. Hence is called the “ Limit” to which the sum of the series approaches That is, is the limit to which the sum of the series may be made to approach as nearly as we please by taking a sufficient number of terms. The difference between the "Limit" of the series and the actual sum of any number of terms (n) is always represented by the quantity |