PROBLEM VI. The first term, number of terms, and the sum of the series being given, to find the last term. Illustration. — Let 3 be the first term, 9 the number of terms, and 99 the sum of the series. By Problem II. it was shown, that, if the sum of the extremes were multiplied by the number of terms, the product was twice the sum of the series; therefore, if twice the sum of the series be divided by the number of terms, the quotient is the sum of the extremes. If from this we subtract the first term, the remainder is the last term; thus 99 X 2= 198; 198 · 9 = 22; 22—3= 19, last term. RULE. - Divide twice the sum of the series by the number of terms; from the quotient take the first term, and the remainder will be the last term. 15. A merchant being indebted to 22 creditors $ 528, ordered his clerk to pay the first $ 3, and the rest increasing in arithmetical progression. What is the difference of the payments, and the last payment ? Ans. Difference 2 ; last payment, $ 45. Section LXIV. GEOMETRICAL SERIES, OR SERIES BY QUOTIENT. If there be three or more numbers, and if there be the same quotient when the second is divided by the first, and the third divided by the second, and the fourth divided by the third, &c., those numbers are in geometrical progression. If the series increase, the quotient is more than unity; if it decrease, it is less than unity. The following series are examples of this kind : 2, 6, 18, 54, 162, 486. 64, 32, 16, 8, 4, 2. The former is called an ascending series, and the latter a descending series. In the first, the quotient is 3, and is called the ratio ; in the second, it is 1. The first and last terms of a series are called extremes, and the other terms means. PROBLE One of the extremes, the ratio, and the number of terme being given, to find the other extreme. Let the first term be 3, the ratio 2, and the number of terms 8, to find the last term. Illustration. — It is evident, that, if we multiply the first term by the ratio, the product will be the second term; and, if we multiply the second term by the ratio, the product will be the third term; and, in this manner, we may carry the series to any desirable extent. By examining the following series, we find that 3, carried to the 8th term, is 384 ; thus, (1.) (2.) (3.) (4.) (5.) (6.) (7.) (8.) 3, 6, 12, 24, 48, 96, 192, 384, ascending series. But the factors of 384 are 3, 2, 2, 2, 2, 2, 2, and 2 ; therefore the continued product of these numbers will produce 384. But, by multiplying 2 by itself six times, and that product by 3, is the same as raising the ratio, 2, to the seventh power, and then multiplying that power by the first term. Hence the following Rule. — Raise the ratio to a power whose index is equal to the number of terms less one; then multiply this power by the first term, and the product is the last term, or other extreme. Illustration. — In the above question the number of terms is 8; we therefore raise 2, the ratio, to the seventh power, it being one less than 8; thus, 2 x 2 x 2 x 2 X 2 X 2 X 2= 128. We then multiply this number by 3, the first term, and the product is the last term; thus, 128 x 3 = 384, last term. The above rule will apply in a descending series. Let the following numbers be a geometrical descending series : — 384, 192, 96, 48, 24, 12, 6, 3, descending series. Let the first term be 384, the number of terms 8, and the ratio , to find the other extreme. By the above rule, we raise the ratio, , to the seventh power, it being one less than the number of terms, 8. Thus, XXX 1 XXX= do. We then multiply this power by the first term; thus, to X 394 = = 3, the extreme required. But as the last term, or any term near the last, is very tedious to be found by continual multiplication, it will often be necessary, in order to ascertain it, to have a series of numbers in arithmetical proportion, called indices or exponents, begin. ning either with a cipher or a unit, whose common difference is one. When the first term of the series and ratio are equal, the indices must begin with a unit, and in this case the prod. uct of any two terms is equal to that term signified by the sum of their indices. S 1, 2, 3, 4, 5, 6, &c., indices or arithmetical series. muss 2, 4, 8, 16, 32, 64, &c., geometrical series. Now 6 + 6 = 12 = the index of the twelfth term, and 64 X 64 = 4096 = the twelfth term. But when the first term of the series and the ratio are different, the indices must begin with a cipher, and the sum of the indices made choice of must be one less than the number of terms given in the question; because 1 in the indices stands over the second term, and 2 in the indices over the third term, &c. And, in this case, the product of any two terms, divided by the first, is equal to that term beyond the first signified by the sum of their indices. 50, 1, 2, 3, 4, 5, 6, &c., indices. '{1, 3, 9, 27, 81, 243, 729, &c., geometrical series. Here, 6+5=11, the index of the 12th term. 729 X 243=177147, the 12th term, because the first term of the series and ratio are different, by which means a cipher stands over the first term. Thus, by the help of these indices, and a few of the first terms in any geometrical series, any term whose distance from the first term is assigned, though it were ever so remote, may be obtained without producing all the terms. 1. If the first term be 4, the ratio 4, and the number of terms 9, what is the last term ? OPERATION. 4. 16. 64. 256 X 256 = 65536 = power of the ratio, whose exponent is less by 1 than the number of terms. 65536 x 4, the first term = 262144 = last term. Or, 4 X 4 = 262144 = last term, as before. 2 If the first term be 262144, the ratio 1, and the number of te. ms 9, what is the last term? Ans. 4. (+)8 =rstyti 65t56 X 262144 = 229 =4, the last term. 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. $ 574.34,91172913250116264106332310802645846357252196069357387776. 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 = second series. 1, 3, 9, 27, 81, = first series. 0, 0, 0, 0, 243 – 1= 242, 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 242 ; 2= 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 3 X 3 X 3 X 3 X 3 X 10 = 7290, last term. 7290 X 3 = 21870 ; 21870 -10= 21860; 21860 ; 3 - 1 = 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 ? Ans. 484. OPERATION. 3 X 3 X 3 X 3 X 3 = 243; 243—1= 242; 242 -3—1= 121 ; 121 X 4 = 484 Ans. 12. If the first term be 6, the ratio , and the number of terms 4, what is the sum of the series? Ans. 9123 OPERATION. X X X = 265; 1=635; 625 - = 68. 1= { =};$25 = *=629 X = 184*=.*; *** Xi = 43=925 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 |