in (a+x)=x9(10+80)=405. Ans. 4. Given a=1, n=100, and d=5, to find ≈ in an ascending series. a+d(n−1)=1+5×99=496. Ans. 5. Given a=1, n=100, and d=2, to find z in a descending series. Ans. 197. 6. Given a 245, n=13, and d=10, to find z in a descending series Ans. 125. 7. Given a 49, n=50, and d = , to find in a descending Ans. 682. series. Note.-If n, d, and z were given to find a in either series, it is manifest that, by reversing the series, a may be substituted for z, and z for a, since the first term of any ascending series is the last term of the same series descending. By reversing the series, it will be, a=125, n=13, and d=10, to find x in an ascending series. Buta+d(n-1)=125+10(13-1)-245. Ans. 9. Given n=7, d=3, and z=90, in an ascending series, to find a. Ans. 72. 10. It is required to find four numbers in arithmetical progression, such that their common difference may be 4, and their continued product 176985. Ans. 15, 19, 23, and 27. 11. The sum of four numbers in arithmetical progression is 56, and the sum of their squares is 864; what are the numbers? Ans. 8, 12, 16 and 20 Promiscuous Questions. 1. What is the sum of a million terms of the series 1, 2, 3, &c.? Ans. 500,000,500,000. 2. What is the sum of 500 terms of the series 1, 3, 5, &c. ?* Ans. 250,000. 3. How often does the hammer of a common clock, keeping correct time, strike the bell in 30 complete days? Ans. 4680 times. 4. If the series 1, 4, 7, &c. be carried to 91 terms, how much will the last term exceed the third? Ans. 269. 5. What must be paid for 80 trees, the price of the first being fixed at 5s. and of the last at 181. all the intermediate prices being in arithmetical progression ? Ans. 7301. 6. What is the number of acres in an estate, which, if sold on the principles of arithmetical progression, 6d. being given for the first acre, and 9d. for the second, the last acre would be valued at 1001. ? Ans. 7999. 7. What sum will be just sufficient to purchase the estate i Question 6th, with all its conditions? Ans. 400,0491. 19s. gd. 8. If, on the 1st of January, 200 copies of a new work be isposed of, and the sale decrease in arithmetical progression to the -nd of the month, in such manner that on the 31st only 20 copies are sold, what is the common difference of the terms Ans. 6. 9. Let 10 and 70 be the two extremes of an arithmetical series, and three the common difference. Required the number of terms, and the sum of the series. Ans. (n) = 21; (s) = 840. 10. What debt can be discharged in a year, by weekly payments in arithmetical progression; the first payment 1s., the last 5.3s. and what is the difference between each payment ? Sum......£135 { { 4 0. 2 C 11. Required the rule for solving the following question :-A person takes out of his pocket, at ten several times, so many different numbers of pounds, every one exceeding the former by 2; the last was 23. Required the sum taken at first. 12. Let the last term of an increasing arithmetical series be 19, the common difference 2, and the number of terms 9. Required the sum of the series. Ans. 99. 12 Find the last term of an arithmetical series; the sum being 99, the number of terms 9, and the first term 3. Ans. 19. The sum of the series 1, 3, 5, &c = n2. 14. Required three numbers in arithmetical progression; the sum of their squares being 1232, and the square of the mean exceeding the product of the extremes by 16. Ans. 16, 20, and 24. 15. Two men, A and B, set out at the same time; A travels 8 miles a-day; and B travels the first day 1 mile, the second 2, the third 3, &c. In how many days will B overtake A ? Ans. 15. 16. There is a number, consisting of three places, whose digits are in arithmetical progression; if it be divided by the sum of its digits, the quotient will be 26; and if 198 be added to the number, the digits will be inverted. What is the number? Ans. 234. 17. If the sum of six numbers in arithmetical progression be 48; and the product of the common difference, multiplied into the least term, be equal to the number of terms; required the common difference and terms. Ans. d=2; And 3, 5, 7, 9, 11, and 13 the terms. 18. In a 100 yards taken in a straight line a stone is placed at the end of every yard. What is the least distance that a person must travel from a basket placed at the beginning of the first yard to pick up each stone separately and return with it to the basket. Ans. 5050 yards. 19. To find the 100th term of the senses 7 x 10 x 13 x &c. Ans. 304. 20. The sum of four whole numbers in arithmetical progression is 10, and the sum of their reciprocals. Required the terms. Ans. 1, 2, 3, 4. Let 2y=common difference, and x-3y=the first or last term. 21. Let the product of five numbers in arithmetical progression be 120, and their sum 15. Required the numbers. Ans. 1, 2, 3, 4, 5. Geometrical Progression. (126.) Definition.-If a series of quantities increase, or decrease, by the continual multiplication or division of the same quantity, then those quantities are said to be in Geometrical progression. G Thus the numbers 1, 2, 4, 8, 16, &c. which increase by the continual multiplication of 2); and the numbers, 1,,,, &c. (which decrease by the continual division of, or multiplication of 4), are in Geometrical progression. And thus it appears that each successive term of a geometrical progression may be considered as formed by multiplying the ratio into the last term; for multiplying by a fraction whose numerator is unity is the same as dividing by the denominator. The general rules are as follows. (127.) Let a denote the first term of a geometrical progression ; r the ratio, or common multiplier; n the number of terms; and s the sum of the series; Then will a+ar+ar2+ar3+...ar"-1=s be a general expression for every geometrical progression where the exponents differ b unity. Multiply both sides of this equation by r and we have ar+ar2 + ar3 +... ar"-1+ar"=rs. From this equation subtract the former and there will remain -a+ar"=rs-s that is a(rn-1)=s(r-1); whence s= a(r"—1) Hence, if any three of the quantities, a, r, n, s, be given, the other may be found from this equation. COROL. If r be less than 1, and n infinite will vanish, therefore the last equation in this case will become s(r—1)=—a, or by a changing the signs of all the terms s(1-r) = a .'. s= ——, and 3 a("=1 a-ar" 310 3(310-210) 8 (59049-1024) = = = 2 1- 310 H 59049 3. Find three numbers in geometrical progression, such, that their sum shall be equal to 7, and the sum of their squares to 21. Let the numbers be x, y, and z; then by question x+y+z=7, x2+y2+x2=21, And by geometrical progression : y :: y : z .'. y2=xz, (A). by the first condition x+x=7-y, .. x+2xx+x=49-14y+y, (B) From this equation subtract twice equation A and there will remain 14y=28, whence y=2; ..x+x=5 consequently x2+2xx+x2=25, and x2+22=17 consequently 2x3 +2x=34, subtract the upper from the lower x-32xx+x=9, extract the square root but x x=3 x+x=5 Whence will be found x=4 and z=1; .. the three numbers are 1, 2, 4. 4. Given a=3, r=5, and n=4, to find s. 468 Ans. 5 The first term of a geometrical series is 2, and the ratio 3; required the sumn of 20 terms of the series. 3486784400. Ans. 6. The sum of ten terms of the geometrical series 1. Required the sum of the series,,, &c. continued for ever. Ans. 1. 2. What is the sum of 99 terms of the series 1, 2, 4, 8, &c. |