OPERATION. 558. To find the number of terms, the extremes and common difference being given. Ex. 1. If the extremes of an arithmetical series are 3 and 19, and the common difference is 2, what is the number of terms ? Ans. 9. It is evident, that, if the difference of the 19 3 extremes be divided by the common dif+1=9. ference, the result will be the number of 2 common differences; thus 19 - 3 16 ; 16 ; 2= 8. Then, as the number of terms must be 1 more than the number of common differences, 8+ 1= 9 is the number of terms in the series. RULE. Divide the difference of the extremes by the common difference, and the quotient increased by 1 will be the required number of terms. EXAMPLES. 2. A man going a journey travelled the first day 7 miles, the last day 51 miles, and each day increased his journey by 4 miles. How many days did he travel ? Ans. 12. 3. In what time can a debt be discharged, supposing the first week's payment to be $1, and the payment of every succeeding week to increase by $ 2, till the last payment shall be $ 103 ? Ans. 52 weeks. OPERATION. 559. To find the sum of all the terms, the extremes and number of terms being given. Ex. 1. The extremes of an arithmetical series are 3 and 19, and the number of terms 9. Required the sum of the series. Ans. 99. In an arithmetical 3 + 19 series the sum of the X 9= 99, Ans. extremes is equal to 2 the sum of two terms Or, 3 + 19 = 22 ; 22 X 44 = 99, Ans. 22 ; 22 X 44 = 99, Ans. that are equally dis tant from them, or to double the middle term, if the number of terms be odd. Thus, in the series, 3, 5, 7, 9, 11, 13, 15, 17, 19, the sum of 3 and 19 is equal to the sum of 5 and 17, or of 7 and 15, and is double the middle term, 11. The reason of this is evident, since 5 and 7 exceed the less ex.. treme by the same quantities by which 17 and 15 are respectively less than the other extreme. Hence, in this latter series, it is evident that, if each term were made 11, half the sum of the extremes, the sum of the whole would remain the same; therefore the sum of the series must equal half the sum of the extremes multiplied by the number of terms, or the sum of the extremes multiplied by hali' the number of terms. RULE. Muliiply half the sum of the extremes by the number of terms. Or, Multiply the sum of the extremes by half the number of terms. EXAMPLES. 2. If the least term of a series of numbers in arithmetical progression be 4, the greatest 100, and the number of terms 17, what is the the sum of the terms ? Ans. 884. 3. Suppose a number of stones were laid a rod distant from each other, for thirty miles, the first stone being a rod from a basket. What distance will that man travel who gathers them up singly, returning with them one by one to the basket ? Ans. 288090 miles 2 rods. 560. To find the sum of the terms, the extremes and common difference being given. Ex. 1. If the two extremes are 3 and 19, and the common difference is 2, what is the sum of the series? It has been shown (Art. 558) that, 19 - 3 if the difference of the extremes be +1=9; divided by the common difference, 2 the quotient will be the number 19 +3 of terms less one. Therefore the X 9 99, Ans. number of terms less one will be 2 8; and 8+1 will equal the number of terms. It has also been shown (Art. 559) that, if the number of terms be multiplied by the sum of the extremes, and the product divided by 2, the quotient will be the sum of the series; therefore 19 + 3 99, the answer required. RULE. - Divide the difference of the extremes by the common difference, and to the quotient add 1; by this sum multiply half the sum of the extremes, and the product will be the sum required. EXAMPLES. 2. If the extremes are 3 and 45, and the common difference 2, what is the sum of the series? Ans. 528. 3. A owes B a certain suin, to be discharged in a year, by OPERATION. 19-3 2 X 9 2 paying 6 cents the first week, 18 cents the second week, and thus to increase every week by 12 cents, till the last payment should be $ 6.18. What is the debt ? Ans. $162.24. OPERATION. [Ans. 561. To find one of the extremes, when the other extreme and the number and sum of the terms are given. Ex. 1. If 3 be the first term of a series, 9 the number of terms, and 99 the sum of the series, what is the last term ? It has been shown (Art. 559) that, if the 99 X 2 sum of the extremes be multiplied by the 3=19, 9 number of terms, the product will be twice the sum of the series; therefore, if twice the sum of the series be divided by the number of terms, the quotient will be the sum of the extremes. If from this we subtract the given extreme, the remainder must be the other extreme. RULE. — Divide twice the sum of the series by the number of terms ; from the quotient take the given term, and the remainder will be the term required. EXAMPLES. 2. The sum of a series of ten thousand even numbers is 100010000, and the last term of the series is 20000. Required the first term. Ans. 2. 3. A merchant, being indebted to 22 creditors $528, ordered his clerk to pay the first $3, and the rest sums increasing in arithmetical progression. What is the difference of the payments, and the last payment? Ans. Difference 2 ; last payment $ 45. 562. To find any number of arithmetical means, the extremes and the number of terms being given. Ex. 1. If the first term of an arithmetical series is 1. the last term 99, and the number of terms 8, what are the second and seventh terms of the series? Ans. The second term, 15 ; the seventh, 85. We find the common difference, 14, as in Art. 557, the first term, 1, plus the common 99 1 difference, 14, gives 15 for 8 1 the second term, and the last term, 99, minus the coinmon 1+ 14 = 15; 99 - 14 = 85. difference, 14, gives the seventh term. OPERATION. 14; RULE. — Find the common difference, which, added to the less extreme, or subtracted from the greater, will give one mean. From that mean derive others in the same way, till those required are found. EXAMPLES. 2. The extremes of a series are 4 and 49, and the number of terms 6. Required the middle two terms. Ans. 22 and 31. 3. Insert five arithmetical means between 20 and 30. Ans. 213, 234, 25, 263, and 281. GEOMETRICAL PROGRESSION. 563. Geometrical Progression, or progression by quotients, is a series of numbers that increase or decrease by a constant multiplier or divisor, called the common ratio. The series is an ascending one, when each term after the first increases by a constant ratio ; and a descending one, when each term after the first decreases by a constant ratio. Thus 2, 6, 18, 54, 162, 486 is an ascending geometrical series; and 64, 32, 16, 8, 4, 2 is a descending geometrical series. Of the former, 3 is the common ratio, and of the lat ter, 2. 564. In geometrical progression the first term, the last term, the number of terms, the common ratio, and the sum of the terms are so related to each other, that, any three of these being given, the other two may be readily determined. 565. To find any proposed term, one of the extremes, the ratio, and the number of terms being given. Ex. 1. If the first term of a geometrical series be 3, the ratio 2, and the number of terms 8, what is the last term ? Ans. 384. It is evident that the successive 3 x 2 = 3 X 128 = 384. terms are the result of repeated multiplications by the ratio ; thus the second term must be the product of the first term by the ratio, the third term the product of the second term by the ratio, and so The eighth, or last term, therefore, must be the result of seven OPERATION on. such multiplications, or the product of the first term, 3, by 27, or 3 x 128 384. If the last term had been given and the first required, the process would evidently have been by division, since every less term is the result of a division of the term next larger by the ratio. RULE. — Raise the ratio to the power whose index is one less than the number of terms ; by which multiply the least term to find the greatest, or divide the greatest to find the least. Note 1. – When the ratio requires to be raised to a high power, the process may be abridged, as in Art 516. NOTE 2. — The rule may be applied in computing compound interest, the principal being the first term, the amount of one dollar for one year the ratio, the time, in years, one less than the number of terms, and the amount the last term. EXAMPLES. Ans. To 2. If the first term be 5, and the ratio 3, what is the seventh term ? Ans. 3645. 3. If the series be 72, 24, 8, &c., and the number of terms 6, what is the last term ? 4. If the larger extreme be 885735, the ratio f and the number of terms 12, what are the tenth and the eleventh terms ? Ans. 45 and 15. 5. If the seventh term is 5, and the ratio }, what is the first term ? Ans. 3645. 6. If the first term is 50, the ratio 1.06, and the number of terms 5, what is the last term ? Ans. 63.123848. 7. 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. 8. What is the amount of $ 160.00 at compound interest for 6 years? Ans. $ 226.96305796096. 9. What is the amount of $ 300.00 at compound interest at 5 per cent. for 8 years ? Ans. $ 443.236+. 10. What is the amount of $ 100.00 at compound interest at 6 per cent. for 30 years ? Ans. $ 574.349117291325011626410633231080264584635 7252196069357387776. 566. To find the sum of a series, the first term, the ratio, and the number of terms being given. |