4. The length of my garden is 94 feet; now if eggs be laid along the pavement 1 foot asunder, and be fetched up singly to a basket, removed one foot from the last : How much ground must he traverse that does it * Answ. I mile, 5 furlongs, 21 poles, 3; feet English, 5. A merchant hires a clerk by covenant 14 years, to give him 51. the first year, and raise his salary 40s, a year ouring the term : The question is, to discover how much he paid him one year with another on an average 2 - Answ. 181. 6. Supposing a press gang haying a warrant to press for 30 days, press the first day 300 men, and every succeedin day 10 more than the former ; how many men will they raise in the 30 days? Answ. 1335. By these problems likewise may the questions relating to annuities in arrear be more readily solved : page 226. For the several yearly or half yearly interests form an Arithmetical Progression, of which the last interest due may be taken as the first term and common difference, and the number of years or half years, &c. iess 1, the Trumber of terms, because there is no interest due upon the last payment, as being only now due. Example, If an annuity of 70l. be forborn 5 years, what will be due at the end of that term for Principal and interest thereof, interest being computed for every annuity from the time it became due, at 5 #' cent. #' annum simple interest A l. 3. *13 10 a year's int. 35 0 amount of interest. 70×3-350 0 amount of the an nuity, Answ. 385 O - * Thus let all the questions in page 227 be solved, and likewise the following, If 70l. annuity, payable by quarterly payments, were unpaid 5 years, what will it amount to in that time, simple interest being computed at 5 #2 cent. Answ. 3911. 11s. 3d. C H A P. IV. GEOMETRICAL PROGRESSION. HEN a Rank or Series of Numbers do either encrease by one common Multiplier, or decrease by one common Divisor, they are said to be in Geometrical Progression, or Geometrical Proportion continued. Asłż 4, 8, 16, 32, 64, here 2 is a com. multiplier, 64, 32, 16, 8, 4, 2, - 2 is a com. divisor, 2, 6, 18, 54, 162, 3 common multiplier, 162, 54, 18, 6, 2, — 3 common divisor, The common Multiplier or Divisor is commonly termed the Ratio of the Progression. - Proposition 1. Any three numbers in geometrical progression will form an analogy, by making the consequent of the former ratio the antecedent of the latter. As the numbers 2, 4, 8, will form this analogy t-4:4:8. Cor. 1. If three numbers be in geometrical progression, the product of the extremes multiplied into each other is equal to the square of the mean. Cor. 2. If the product of two numbers be equal to the square of a third, these numbers are in geometrical progression. - Problem I, Having two numbers to find a mean proportional between them. Multiply the two numbers into each other, and extract the square root of the product. Find a mean proportional between 4 aud 9 Answ. 6. What is the mean proportional between 4 and 64 ° Answ. 16. Proposition 2. Any four numbers in geometrical progression will form an analogy or proportion. As 2, 4, 8, 16, 2 : 4 :: 8:16, which is manifest. Cor. Therefore, if four numbers be in geometrical progression, the product of the means will be equal to the product of the extremes. Proposition Proposition 3. In any geometrical progression the product of the two extremes, is equal to the product of any two terms equally distant from the two extremes. * 3, 6, 12, 24, 48, 96, 3, 6, 12, 24, 48, 96, 192, As 3×96=23S and 6 x 48 =288, &c. If over a geometrical progression beginning with unity, we place 1 over the second term and so proceed orderly according to the natural progression of numbers, viz. O 1 2 3 4 5 6 7 8 Indices. 1, 2, 4, 8, 16, 32, 64, 128, 256, Powers, The numbers 1, 2, 3, &c. will express what power of the ratic the term is over which it stands, and are therefore called indices or exponents of the power. If the first term given be the ratio, then every succeeding term is the same power of the ratio as the order of its place, or the index of the power will denote both the power and order of the place, for i must be placed over the first term, 2 over the second, &c. 1 2 3 4 5 6 7 Indices. 2, 4, 8, 16, 32, 64, 128, Powers. 2 Proposition 4. In any geometrical progression beginning with unity, (if the indices be supposed placed over the terms of the progression) there will be this coherence or relation be. tween the powers and their indices, viz. The sum of the indices of any two powers, or terms of the progression will be the index of the product of the said two terms, and if the index of any term be doubled, its doubie will be the index of the square of said term in the said progression. O 1 2 3 4 5 6 7 8 1, 2, 4, 8, 16, 32 64, 128, 256, Powers. As 3+5=8 and 8 x 32=256, Likewsse 4+4–8 and 16 × 16=256, &c. Proposition 5. ... In a geometrical progression not proceeding from unity, if any term be squared, and the square be divided by the first or least terru, the quotient gives a term of the same, progression doubiy distant from the first. C. c 2 O 1 0 1 2 3 4 5 6 7 8 In any geometrical progression not proceeding from unity, if any two terms be multiplied together, and the product divided by 4 the first or least term, the quotient will be equal to that term denoted by the sum of the exponents of the other two. Proposition 7. In any geometrical progression, as any one of the antecedents is to its consequent, so is the sum of all the antecedents to the sum of all the consequents. 2, 4, 8, 16, 32, 64, &c. 2 : 4:: 2+4+8+16+ 32(62) +8+16+32-H 64(124) * Problem II. To continue a geometical progression upwards or downwards. 1. Upwards; divide any consequent given by its antecedent, and the quotient will be the ratio, whereby multiply the consequent and the product will be the next term, which being again multiplied by the ratio, will produce a new term, and so on, and it is continued downwards by dividing each greater term by the ratio. Problem III. To find any assigned term of a geometrical progression proceeding from unity without producing all the terms. Continue the progression from unity to the sixth term whose index is 5, square this sixth term, and it produces that whose index is 19, which being likewise squared, its square will be the term whose index is 20; and from those terms we may easily find all others whose indexes are decades or even tens, viz. 20+10=36; 20+20–40, and 40 + \ 40+ 10–50, &c. and from them any other whatever. For the units are either greater or not greater than 5; if not greater, multiply the term last found by the term whose index denotes the distance of the assigned term from that last found ; and if greater, multiply first by the term under 5, and then by the surplus of the units or index of the assigned term above 5. Application. Let it be required to find the 43rd term of a geometrieal progression, beginning with unity whose ratio is 2 o The progression being con5 6 7 tinued to the 5th place 32, , 32, 64, 128, whose index is 5, 32 × 32= - 1024, which will be doubly distant from unity, viz. the 11th term of the progression, whose index is 10. Again, 1024x 1024 produces 1048576 the 21st, which being again multiplied by itself, will produce that term whose index is 40, viz. the 41st; but it is proposed to find the 43rd, viz. that whose index is 42, and 40+2=42; wherefore multiply the last found number which is 1099511627776 by 4, the number whose index is 2, and so we get the 43rd term required, 43980465 l 1104. Problem IV. To find any assigned term of a geometrical progression not beginning with unity, without producing all the terms. Proceed directly as in the last problem, only observe to divide every product by the first term. Application. Let it be required to find the 26th term of a geometrical progression whose first term is 2 and ratio 3 ° I continue the progression to the sixth place, which being 0 1 2 3 4 5 squared, and the product divided 2, 6, 18, 54, 162, 486, by the first term 2, the result is 118098 whose index is 10; this being again squared, and the product divided by 2, produces 6,973,568,802, whose index is 20, and this last being multiplied by 486 and divided by 2, produces 1,694,577,218,886, whose index is 25, which is the 26th term. Problem V. To find the sum of any geometrical progression. |