123. To shew how the 17th, 18th, 19th, and 20th theorems are derived. Since Z=A+R.n-1 (th. 2.) therefore n-1= Z-A n= +1 (THEOR. 17.) and because R=L.s-a-L.s—z (th. 4.) substitute this value for R in theor. 17. and n= Z-A +1 (THEOR. 18.) again, for Z in theor. 17. sub Z-A substitute its value from theor. 5. and n= +1=) R EXAMPLES.-1. Given the ratio 2, the number of terms 6, and the last term 96, of a geometrical progression, to find the first term, and the sum of the terms? Here r=2, n=6, z=96, whence (theor. 1.) a= 1 1.9822712 2. Given the ratio 2, the number of terms 6, and the sum of the terms 189, to find the first and last terms? r-1.s Here r=2, n=6, s=189, and (theor. 9.) a= By Logarithms. 0.0000000 | Lr—1+R.n—1= 1.5051500 2.2764617 1.7993405 L.r−1+R.n—1+S= 3.7816117 2.2764617 | +S= 3. Given the first term 3, the ratio 2, and the last term 96, to find the number, and sum of the terms? Here a=3, r=2, z=96, and (theor. 7.) rz-a 2 × 96-3 - 1 L.189=2.2764617 -Lr-1=L.1= 0.0000000 S= 2.2764617 whence s=189. .... R=0.3010300) 1.5051500(5 therefore n=5+1=6, theor 17. 4. Given the first term 4, the ratio 3, and the sum of the terms 484, to find the last term, and number of terms? Here a=4, r=3, s=484, and (theor. 6.) z= n the first term 2, last term 2048, and sum of the O, to find the ratio, and number of terms? a=2, z=2048, s=2730, and (theor. 4.) r= = 2 2728 -L.s-z L.682=2.8337844 L.s-a-L.s-z=0.6020600 therefore .6020600)3.0103000(5 Ans. a= =en r=4, n=6, and s=2730, to find a and z. 048. _ven r=2, n=6, and z=96, to find a and s.. Ans. a=3, ven the ratio 5, last term 12500, and sum of the terms to find the first term, and number of terms. Ans. a=4, iven a=4, n=6, and z=12500, to find r and s. Answer =15624. Given r=3, n=4, and z=81, to find a and s. PROBLEMS IN GEOMETRICAL PROGRESSION. Of three numbers in geometrical progression, the difference first and second is 4, and of the second and third 12; red the numbers? Let x, y, and z, be the numbers. :y: Then y—x=4, or x=y—4; z—y=12, or z=y+12. Wherefore since by the problem xy:: y: z, by substituting alues of x and z in this analogy, we shall have y —4 : y :: 12; wherefore, (by multiplying extremes and means,) y-4 12=) y2+8y—48=y2, or 8y=48; wherefore y=6, x=2, 18. 2. The product of three numbers in geometrical progression is 1000, and the sum of the first and last 25; required the numbers? Let x, y, and z, be the numbers; then since xy::y:z, we have xz=y, (Art. 120. Note,) and (xyz=xz.y=) y3=1000, whence y=10; also xz=(y2=) 100, and by the problem x+z= 25: from the square of this equation subtract four times the preceding, and x2 — 2xz+z2=225: extract the square root of this, and x-z=15; add this to, and subtract it from, the equation x+z=25, and 2x=40, or x=20, also 2 z=10, or z=5; whence 5, 10, and 20, are the numbers. 3. To find any number of mean proportionals between two given numbers a and b. Let n-2=the number of mean proportionals, then will n= the number of terms in the progression: also let r=the ratio, then (theor. 3. Geom. Prog.) r= b; and by logarithms, log. b—log, a a ÷n-1=log. r; whence r being found, if the less extreme be continually multiplied, or the greater divided, by r, the results will be the mean proportionals required. EXAMPLES.-1. To find two mean proportionals between 12 and 4116. Here a=12, b=4116, n=4, and r= 4116 =343+=)7; whence 12×7=84, the first mean, and 84×7=588, the second mean. 2. To find four mean proportionals between 2 and 486. Ans. 6, 18, 54, and 162. 3. To find five mean proportionals between 1 and 64. 4. There are four numbers in geometrical progression; the sum of the extremes is 9, and the sum of the cubes of the means 72; what are the numbers? Let x, y, u, and z, be the numbers. Then by the problem, x+z=9, or x=9-z. xy::u: z, or xz=uy, whence xz=(9—z.z=) 9z—2°. |