EXERCISES IN GEOMETRICAL PROGRESSION. 1. Here a=1, r=2, and n=10; therefore Theorem 9, 1=29—25 × 21=32 × 16=512. Theorem 6, s= 210-1 =25x25-1-32 x 32-1-1023. 2. Here a=1, 1=128, and s=255; therefore substituting these values in Theorem 4th, we have may 3. Here a=1, r, and n= infinity; therefore be considered as =0. Hence (4) equation (Art. 96), gives 4. Here a=1, r=3, and n= infinity; therefore 6. Here a=1 farthing, r=4, and n=12; therefore 412-1 (44)3-1 (256)3-1 Theorem 6 gives s= = = 4-1 3 3 = 16777215 farthings, L.5825, 8s. 54d. 7. Let the first term be represented by x, and the common ratio by r, then xr will be the second term, and xr2 the third; hence 1x+xr+xr2=52, 2 Given equations. 3 5x+5x=2x+2xr2, by mult. ext. and means of (2). 4 | 5+5r=2+2r2, dividing by x. 5 2r2-5r-3, transp. and changing sides. 8 57 =3, or 4 Substituting the value 3 instead of r in (1), we obtain 13x=52, ..x=4; hence we have the numbers sought, 4, 12, and 36. be If the other value of r be substituted, the numbers will the question; but one of them being minus, they cannot properly be considered as included in the enunciation of the question. then 8. Let the notation be the same as in the last exercise; By the conditions of the question. 1x+xr15, 1+r2 (121.) 1. By Formula (1), Log. 1000=3·0000000000 By Table 12; 10 Log. 1·05 =0·2118929910 Ans. L.1628, 17s. 93d. =1628.89–3·2118929910 2. By Formula (2), Log. 700 =2.845098 H 11 Log. 1.04 0.1873667323 Ans. L.454, 14s. 1d. =454.706=2·657731 5. Apply Formula (4). This question is the same as,In what time will one pound become two, at 5 per cent? Log. 2-Log. 1 •301030 hence === Log. 1.05 ⚫021189 =14.2 years. 8. Here the amount must first be found (1). And the value of R will be found by (Art. 121); we have then, Log. 100+12, Log. 104-Log. 160-103-L.160, 2s. 03d. Log. 100+24, Log. 1.02 Log. 160-844-L.160, 16s. 10 d. Log. 100+48, Log. 1.01 Log. 161-223=L.161, 4s. 5d. Subtracting the principal, L.100, from each of the above, we obtain the compound interest sought. (127.) 1. Apply Formula, (Art. 125), and 2000, (1—(1+r)") r =623111; hence 2000 ×·623111=1246·222—L.1246, 4s. 51d. 2. Apply the last formula in (Art. 125), viz, p= 100 100 hence the value sought is =100 x =L.2000. 5 a 3. This is the same as finding the present value of an annuity of L.50 for 40 years, at 24 per cent, to which ap 1 ply formula, (Art. 125), viz. p=;(1— (1+r)" a (1+r)n), =2000 (1-372431)=2000 x 627569-L.1255, 2s. 94d. 4. This is the same as finding the value of a perpetual annuity of L.50 per annum at one-half of 5 per cent; and hence we have, by substituting these values in the last formula, (Art. 125), 5. By (Art. 126), and substituting the proper values of t, n, and r, we have 6. Here p= 6761.59 11.4674 (1.05)4-1 50 6.76156 5000 =589-636-L.589, 12s. 8d. 5 7. By (Art. 127), calling the amount a, we have ..+1=(1+r)", by mult. by and transp. 8. In Formula, (Art. 127), put A=24, r='05, or g'ō, A (1+r)=1·05, and n=20, then=480, and {(1+r)20—1} =1.6533, which being multiplied by 480, gives 793-584; therefore the amount sought is L.793, 11s. 8d. NOTE. The exercises in compound interest may be easily solved by Tables 7th and 8th of this work; and those in annuities by Tables 9th and 10th. PROMISCUOUS EXERCISES. Page 78. (1.) Let x be the first part of 11, then 11- -x will be its second part; also let y be the first part of 17, then 17-y will be its second part. (2)—(1) 1|xy=45, By the 2 187-11y-17x+xy=48, question. 3 187-11y-17x=3. 4 11y 184-17x. 184-17x 5y= 11 (3) transp. (4)÷11 (1) by substitution (2.) Let x represent the first part of 30; then 3x will represent the first part of 21; and therefore the second part of 21 will be represented by 21-3x, and the second part of 30 by 30-x; hence by the question, hence (21-3x)2+(30-x)2=585. :.x=6, 30-x-30-6-24, 3x=18, and 21-3x-21-18=3, Values sought. square. (3.) Let x represent the sum left to the first; then since the shares were in continued proportion, if r be put for the common ratio, the second's share will be rx, and the third's r2x. (1)÷(2) (7) transposed (9)× by 4 43+3r+3=7—7p2. 5 10r2+3r=4. 6 (400r2+120r+9=169, by completing square. 720r+3-13, by extracting the root. 8 r, and divided by 20. (2) by substitution 9 x-x=90. 10 3x-360. 11 x 120; hence rx=60, and r2x=30. (4.) Let x represent the greater of the two numbers; then since their product is 45, the other will be by the question we have 45 ; and |