264 GEOMETRICAL SERIES. [SECT. LXIV 3. If the first term be 72, the ratio, and the number of terms 6, what is the last term ? Ans.. 4. If I were to buy 30 oxen, giving 2 cents for the first ox, 4 cents for the second, 8 cents for the third, &c., what would be the price of the last ox? Ans. $10737418.24, 5. If the first term be 5, and the ratio 3, what is the seventh term? Ans. 3645. 6. If the first term be 50, the ratio 1.06, and the number of terms 5, what is the last term? Ans. 63.123848. 7. What is the amount of $160.00 at compound interest for 6 years? Ans. $226.96,305796096. 8. What is the amount of $ 300.00 at compound interest at 5 per cent. for 8 years? Ans. $443.23,6+. 9. What is the amount of $100.00 at 6 per cent. for 30 years? Ans. $574.34,9117291325011626410633231080264584635 7252196069357387776. PROBLEM II. The first term, the ratio, and the number of terms being given, to find the sum of all the terms. In order that the pupil may understand the following rule, we will examine a question analytically. Let the following be a geometrical series, and we wish to obtain its sum: 1, 3, 9, 27, 81. Illustration. By examining this series, we find the first term to be 1, the last term 81, the ratio 3. If we multiply each term of the following series, 1, 3, 9, 27, 81, by 3, the ratio, their product will be 3, 9, 27, 81, 243, and the sum of this last series will be three times as much as the first series. The dif ference, therefore, between these series will be twice as much as the first series. 3, 9, 27, 81, 243 1, 3, 9, 27, 81, 0, 0, 0, 0, 243 second series. 1242, difference of the series. As this difference must be twice the sum of the first series, therefore the sum of the first series must be 2422 121. By examining the above series, we find the terms in both the same, with the exception of the first term in the first series, and the last term in the second series. We have only, then, to subtract the first term in the first series from the last term in the second series, and the remainder is twice the sum of the first series; and half of this being taken gives the sum of the series required. RULE. - Find the other extreme, as before, multiply it by the ratio, and from the product subtract the given extreme. Divide the remainder by the ratio less 1 (unless the ratio be less than a unit, in which case the ratio must be subtracted from 1), and the quotient will be the sum of the series required. See operation, question 10. Or, raise the ratio to a power whose index is equal to the number of terms; from which subtract 1, divide the remainder by the ratio less 1, and the quotient, multiplied by the given extreme, will give the sum of the series. See operation, question 11. But if the ratio be a fraction less than a unit, raise the ratio to a power whose index shall be equal to the number of terms; subtract this power from 1, divide the remainder by the difference between 1 and the ratio, and the quotient, multiplied by the given extreme, will give the sum of the series required. See operation, question 12. 10. If the first term be 10, the ratio 3, and the number of terms 7, what is the sum of the series? Ans. 10930. OPERATION. 3 X 3X 3X 3X 3X3 X 10=7290, last term. 7290 X 321870; 21870 - 1021860; 21860 ÷ 3 =10930 Ans. 11. If the first term be 4, the ratio 3, and the number of terms 5, what is the sum of the series? OPERATION. Ans. 484. 3x3 3 x 3 x 3X 3X 3X 3X 3 243; 243-1242; = 242÷3-1= 121; 121 × 4484 Ans. 12. If the first term be 6, the ratio, and the number of terms 4, what is the sum of the series? OPERATION. 16 Ans. 9,995. ? ×?×? × = 1; 1= 83}; 838 — 62% = 828. 1={; {-}=}; $28÷ 3 = 628 × &=3843 = 728; 72Z Xf=3=99 Ans. 13. How large a debt may be discharged in a year, by pay. ing $1 the first month, $ 10 the second, and so on, in a tenfold proportion, each month? Ans. $111111111111. 14. A gentleman offered a house for sale, on the following terms; that for the first door he should charge 10 cents, for the second 20 cents, for the third 40 cents, and so on in a geo 12 metrical ratio, there being 40 doors. What was the price of the house? Ans. 109951162777.50. 15. If the first term be 50, the ratio 1.06, and the number of terms 4, what is the sum of the series? Ans. 218.7308. 16. A gentleman deposited annually $10 in a bank, from the time his son was born until he was 20 years of age. Required the amount of the deposits at 6 per cent., compound interest, when his son was 21 years old. Ans. $423.92,2+. 17. If the first term be 7, the ratio, and the number of terms 5, what is the sum of the series? Ans. 98 83 256 18. If one mill had been put at interest at the commencement of the Christian era, what would it amount to at compound interest, supposing the principal to have doubled itself every 12 years, January 1, 1837 ? Ans. $11417981541647679048466287755595961091061972.99,2. If this sum was all in dollars, it would take the present inhabitants of the globe more than 1,000,000 years to count it. If it was reduced to its value in pure gold, and was formed into a globe, it would be many million times larger than all the bodies that compose the solar system. PROBLEM III. To find the sum of the second powers of any number of terms, whose roots differ by unity. RULE. Add one to the number of terms, and multiply this sum by the number of terms; then add one to twice the number of terms, and multiply this sum by the former product, and the last product, divided by 6, will give the sum of all the terms. 19. What is the sum of 10 terms of the series 12, 22, 32, 42, 52, 62, 72, 82, 92, 102? 20. What is the sum of 100 terms of the series 12, 22, 32, 42, 52, 62, 72, 82, 92, 102, &c., to 1002 ? Ans. 338350. 21. Purchased 50 lots of land; the first was one rod square, the second was two rods square, the third was three rods square, and so on, the last being 50 rods square. square rods were there in the 50 lots? How many Ans. 42925. 22. Let it be required to find the number of cannon shot in a square pile, whose side is 80. Ans. 173880. NOTE.- A square pile is formed by continued horizontal courses of shot laid one above another, and these courses are squares, whose sides decrease by unity from the bottom of the pile to the top row, which is composed of only one shot. PROBLEM IV. To find the sum of the third power of any number of terms, whose roots differ by unity. RULE. Add one to the number of terms, and multiply this sum by half the number of terms; the square of this product is the sum of all the series. 23. Required the sum of the following series: 133, 23, 33, 43, 53, 63, 73, 83, 93, 103, 113, 123. OPERATION. 12+1=13; 12+2=6; 13×6=78; 78X786084 Ans. 24. I have 10 blocks of marble, each of which is an exact cube. A side of the first cube measures one foot, a side of the second 2 feet, a side of the third 3 feet, and so on to the 10th, whose side measures 10 feet. Required the number of cubical feet in the blocks? Ans. 3025 cubic feet. 25. What is the sum of 50 terms of the series 13, 23, 33, 43, 53, 63, 73, &c., up to 503 ? Ans. 1625625. SECTION LXV. INFINITE SERIES. AN INFINITE SERIES is such as, being continued, would run on ad infinitum; but the nature of its progression is such, that, by having a few of its terms given, the others to any extent may be known. Such are the following series : 1, 2, 4, 8, 16, 32, 64, 125, 25, 5, 1, 6, 25, 125, 128, &c., ad infinitum. 65, &c., ad infinitum. To find the sum of a decreasing series. RULE. Multiply the first term by the ratio, and divide the product by the ratio less 1, and the quotient is the sum of an infinite decreasing series. T 1. What is the sum of the series 4, 1, 4, 7, 4, &c., continued to an infinite number of terms? OPERATION. 4 X 4 = 5 Answer. 2. What is the sum of the series 5, 1, 1, 2, &c., continued to infinity? Ans. 61. 3. If the following series, 8, 4, 49, 33, &c., were carried to infinity, what would be its sum? Ans. 9. 8 4. What is the sum of the following series, carried to infinity: 1,,, 27, 1, &c. ? Ans. 14. 5. What is the sum of the following series, carried to infinity: 11,,, &c.? Ans. 12. 6. If the series, 1, 1, 12, 24, &c., were carried to infinity, what would be its sum ? Ans. 14. SECTION LXVI. DISCOUNT BY COMPOUND INTEREST. 1. What is the present worth of $ 600.00, due 3 years hence, at 6 per cent. compound interest? 3 OPERATION. 1.06) = 1.191016)600.00($ 503.77+ Ans. By analysis. We find the amount of $1 at compound interest for 3 years to be $1.191016; therefore $1 is the present worth of $ 1.191016 due 3 years hence. And if $1 is the present worth of $ 1.191016, the present worth of RULE. Divide the debt by the amount of one dollar for the given time, and the quotient is the present worth, which, if subtracted from the debt, will leave the discount. 2. What is the present worth of $500.00, due 4 years hence, at 6 per cent. compound interest? Ans. $396.04,6+. 3. What is the present worth of $1000.00, due 10 years hence, at 5 per cent. compound interest? Ans. $613.91,3+. 4. What is the discount on $800.00, due 2 years hence, at 6 per cent. compound interest ? Ans. $88.00,3-+-. 5. What is the present worth of $1728, due 5 years hence, Ans. $1291.26. 10 years hence, at 5 Ans. $1423.52. at 6 per cent. compound interest? 6. What is the discount on $3700, due per cent. discount, compound interest? 7. What is the present worth of $7000, due 2 years hence, at 5 per cent. compound interest ? Ans. $63492.10. |