(2) What will it cost in U. S. money to settle a debt of 18270 francs in Paris, exchange being 5.22 francs to a dollar? Solution: Since 5.22 francs = $1, 18270 francs are equal to as many dollars as 5.22 is contained times in 18270, or $3500, Ans. (3) What cost a bill on Berlin for 540 marks, at $.941, brokerage being %? Solution: Since 4 marks are worth $.945, the worth of 540 marks is 540 times of $.945, or $127.58 cost less brokerage. $127.58 +1% 488. The method of finding the face of a foreign bill of exchange is essentially the same as that of domestic bills. (4) What is the face of a bill of exchange on London, bought for $4500, at $4.87 in gold? Solution: Since $4.875 will buy a bill of £1, $4500 will buy as many pounds as $4.875 is contained times in $4500, and $4500 ÷ 4.875 £923.076, or £923. 1 s. 61 d.,. Ans. = EXERCISE CXIV. 5. What is the face of a bill on Dublin, for which $6500 was paid in gold, at $4.86? 6. Find the face of a bill on Geneva, which cost $1500 gold, exchange 5.16. 7. Find the cost of a bill for 1000 francs on Antwerp, at 5.17 fr. to a dollar, gold at 1% premium? 8. What would be the cost of a sight bill on Berlin for 1680 marks, when exchange is quoted at .961? 9. What would be the cost in St. Louis of a sight draft on Hamburg for 3200 marks, when exchange is quoted at .961? 10. How large a sight draft on London can be bought in New York for $1174.20, when sterling exchange is quoted at 4.891? 11. What will a bill on London for £200 12 s., payable in 60 days, cost in New York, when sterling exchange is quoted at 4.85%? 12. What will a sight draft on London for £300 8 s. cost, when sterling exchange is quoted at 4.88? 13. Wishing to pay a bill of £860 15 s. in Liverpool, I buy a bill of exchange at 60 days' sight on London; what does it cost, exchange being at 4.87 and brokerage %? 14. Find the cost of a bill of exchange on Geneva, Switzerland, for 25600 francs, exchange being 5.18 francs to the dollar, and brokerage %. SPECIFIC GRAVITY. 489. Table of specific gravities of a few well-known substances. ARITHMETICAL PROGRESSION. 490. An arithmetical progression or series is a succession of numbers, each of which is greater or less than the preceding number by a constant difference. (1) 2, 5, 8, 11, 14, etc., is an increasing arithmetical progression, in which the common difference is 3. (2) 27, 23, 19, 15, 11, etc., is a decreasing arithmetical progression, in which the common difference is 4. There are five elements in every arithmetical progression, viz.: the first term (a), the last term (7), the number of terms (»), the common difference (d), and the sum of all the terms (s). The first and last terms are the extremes. If any three of these elements are given the other two may be found. 491. To find the last term when the first term, common difference and number of terms are given. (1) If the following increasing progression, 5, 8, 11, 14, 17, etc., were extended until there were 15 terms in all, what would the last term be? Explanation.--5, the first term, in Operation. creased by once 3 gives the second term, 5 + (15 — 1)3 = 47, Ans. by twice 3 gives the third term, by three times 3 gives the fourth term, etc. Hence, 3 increased by fourteen times 3 gives the fifteenth term. Formula: l = a + (n − 1) × d. (2) If the following decreasing progression, 41, 37, 33, 29, 25, etc., were extended until there were 10 terms in all, what would the last term be? Explanation.-41, the first term, de Operation. = 5, Ans. creased by once 4 (the common differ- 41 (10 — 1)4 ence) gives the second term, by twice 4 gives the third term, by three times 4 gives the fourth term. Hence, 41 decreased by nine times 4 RULE.-Multiply the common difference by a number less by 1 than the number of terms; add the product to the first term of an increasing series, subtract it from the first term of a decreasing series, and the result will be the last term. EXERCISE CXV. 3. Find the 30th term of the series 4, 9, 14, 19, etc. 8. Find the 21st term of the series 103, 100, 98, etc. 9. Find the last term of an ascending series, the first term of which is 5, the common difference 3, and the number of terms 51. 10. The first term of a descending series is 200, the common difference 3, and the number of terms 47; what is the last? 492. To find the first term when the last term, number of terms, and common difference are given. By reversing the terms of a series, the last term of an increasing series becomes the first term of a decreasing series, and the last term of a decreasing series becomes the first term of an increasing series. Hence, RULE.-Find the first term of an increasing series as if it were the last term of a decreasing series, and find the first term of a decreasing series as if it were the last term of an increasing series. 11. Find the first term of an increasing arithmetical series, the last term being 191, the number of terms 45, and the common difference 4. 12. The last term of a decreasing arithmetical series is 7, the number of terms 31, and the common difference 5; what is the first term? 13. A man deposited in a savings bank $7 for his son when the boy was one year old, and increased the deposit by $10 at each subsequent birthday until his son was 21 years old; what was the last deposit? 493. To find the sum of all the terms when one of the extremes, the number of terms, and the common difference are given. (14) Find the sum of the eight terms of the following arithmetical progression: 3, 7, 11, 15, 19, 23, 27, 31. Explanation.—The average of any Operation. two terms equally distant from theof (3 + 31) X 8 = 136, Ans. extremes, as 7 and 27, is of (3 + 31), or 17. Hence, the average of all the terms is 17; therefore the sum of the eight terms is 8 X 17, or 136. RULE.-Multiply half the sum of the extremes by the number of terms. NOTE. If either of the extremes is not given, find it as in Art. 491, or 492. 15. If the first term is 5, the last term 62, and the number of terms 20, what is the sum of the terms? 16. Find the sum of the integers 1, 2, 3, 4, 5, etc., carried to 100 terms. 17. Find the sum of the odd numbers from 1 to 101 inclusive, of which there are 51. Find the sum: 18. Of 20 terms of the series 2, 6, 10, 14, etc. 19. Of 37 terms of the series 165, 161, 157, etc. |