100a-art 100 cct(r-1) 100cr-(r-1)×100a ort= ; Also r-1 then tt. By ques. (a-c)XrX(r—1) (r+1)100a-100br 100cr-(r-1)X100a (b-a)XrX(r+1) (a-c)rX(r-1) (b-c) Xa by reduction, (b+c)ar-2brc= (b—c)a, or r — (12696)112-(2×112×96) (b+c)a-2bc 536. If I lend $150 on condition that $12 per annum be paid until the principal and its interest, at 5 per cent,. be satisfied; and that at every payment the interest then due shall be discharged, and the remainder, by which such payment exceeds that interest, applied to reduce the principal, how many years must the said payment continue? See Index. = a+pr= the interest due at Let p= $150, a= 12, r= .05, and n― time required; then p+ pr= the sum due before first payment, p sum due after first payment, and (p-a)r+pr2 second payment, (2p-a)r+pr2: sum due after second payment. Also, (p2a)r+ (2p—a)2 + pr3: the interest due at 3d pay. ment; p-3+(3p—3a)r+(3p—a)r2+pr3: the sum due after third payment. Again, (p-3a)r+(3p—3a)r2+(3p—a)r3+pr1 the interest due at the fourth payment, p-4a+(4p-6a)r+ (6p-4a)2+(4p-a)r3+pr1 = sum due after fourth payment; n(n-1) n(n-1)(n-2) then I shall have p+npr+ -DU n(n-1) n(n-1)(n-2) -ar 2 1.2.3 &c., will be the sum due after the nth payment; which, by the n(n-1),,' &c. = question, is nothing. Then p X 1 + nr + 1.2 537. The duties of certain goods amounted to $2460, out of which a discount of 21 per cent. was allowed, in consideration of prompt payment on the sum actually paid. What did the discount amount to? Put $2460 s, föf=7&=r; 1=t, and d= discount. Then 2460X o 1+10 38. A person has now due to him $320, and at the end of 5 years $96 more will be due from the same debtor; they agree, that the whole shall be discharged at one payment, at that time when the simple interest of the $320 shall be equal to the discount of the $96, both being calculated at 5 per cent. per annum. of payment is required. Let 320 =P, t= = 5, $96 — s, 180 time; then (t—x) × pr— interest of p for t the discount of n for x; then (t—x)pr =tpr s+p-tpr tp t pr 2pr The time x, and xpr or -x=-=- ; by art. 70, case pr r s+p-tpr 2pr ➡4, and then 5 -41, the time required is one year, or, expressed in numbers, 96320-5(320×2) =21, and x=√ { (5×20) +21"} 39. Bought as many books, at 4s. and 6s. apiece, to pay in 9 and 6 months respectively, as cost £33 2s, but in consideration of prompt payment am allowed the just discount, which is £1 2s.. and which is 9 d. less than the interest of the price of the said two parcels, each being calculated for its respective time. Required the number of each sort of books, and the rate of interest. Let x = the number at 4 shillings, z = the number at 6 shillings, and r the rate of interest; then 4x+6x=662 shillings; rx 9rz then + =22&fs. = sum of the interest, and 50 18rz 400+3r 200 4rx 200+r + =22s. = sum of the discount. By the first equation, 331-3z which substituted for it in the other two, we have tuted for it in the other, and reduced, it is 24r2 — 275r — — 775, from which r = 5, then z = 83, and x=41. Otherwise, Let x the price of that parcel at 4 shillings apiece, and r = the interest of one pound for a year; then r = interest of the price of that parcel at 4s. apiece, and r X (331-x) — the interest of the price of that parcel at 6s. apiece; the sum of both is rx 9937-10rx 993r rx ; also rx discount of 40 3 1+화 331-x xr+ the first parcel, and XrX count of the second parcel; their sum is 40+30r 2+r =22s.£11 by the question, which equation reduced, we have 240r2469r 22 £1 2s.+9d.= £, and this equation reduced, we have x 19860r-911 200r 2. -ᄒᄒ. do; by art. 70, r= , hence x164—£8 4s., and 164 {(4s.) 41 = the number of books at 4s. apiece; whence will be found 83= the number at 6s. apiece. 540. Let as before, art. 137, &c. Put P= principal or present worth, a = the annuity, r➡ rate of interest of one dollar for one year, n = number of years or times, m = the amount; then if p, r, and n are given, to find m. Now the amount of one dollar in one year will equal 1+r, ..pp(1+r), and p(1+r) will amount to p(1+r) in 2 years, and so on; p(1+r)2 will=p(1+r)3 in 3d year; and in general p(1+r) will amount p(1+r)"= the amount of (p) at the end of (n) years or times; therefore p(1+r)" m. In logarithms, P÷nL(1+r) = M, and cor. 1, =M, M-P when p, r, and m are given; cor. 2, when n and m are given, to find p, then P=M―nL(1+r). 541. Given the annuity a, rate per cent. r, and number of years or times➡n, to find m = the amount. Then I have a+a(1+r)+a(1+r2)+a(1+r3), &c., to a(1+r); by art. 3, page 180, will become due in the times 1, 2, 3, 4, a(1+r)" — a and so on; the sum of the series is =m. In Loga n× r rithms: Let nX L.(1+r) B; then A+L(-1) — R=M; Cor. 2. When m, r, and n, are given to find a. See Index. rm or ifn XL((1+r) B)AM+B-L.(-1); (1+r)-1 Cor. 2 when a, m, and r, are given to find n, a(1+r)”=mr+a .. (1+r)"= mr+a and n= " a L(mr+a)—A 1. When a, n, and m are given, to find r, by arts. 86 to 92. a(1+r)" — a (1+r) — 1 (=m, or a(1+r)”—a—m(1+r)—m; then a(1+r)” -m(1+r)+(m − a)==0. Let D `also F = = M-A-N ; e = (n-1) L{2(d—1) + e} +E; then r=f-e. 2 6 n+1 542. If an annuity of $50 forborne 18 years amount to $1342.75 cts., what rate of interest was allowed? Answer is 4 per cent. Here a 50, n = 18, and 1342.75=m, and 8.5 ={(n−1); then f = 0.36078 · 0.31579 1,61508=L.{2(d—1)+e} 2)1,11448, 1.55724=F; Given a, r, and n, to find p, In annuities at compound interest. Let n XL(1+r) = B; ...A + L.(6-1) — R— B = P. (1+r)-1 544. When a, p, and r are given, to find n. L(6-1) A-L(a-pr) ; then p(1+r)2 — (a + {(1+r)—1}× (1+r)* p) (1+r)" + a=0. Let G = ; then rh-k. See Index. 545. In annuities, computed by Comp. Interest, a, p, and rare given, to find m. By my Index, and also at pages 315, '6, "7, that -pr a+mr mp m-p 546. In computing annuities by Compound Interest, p, m, and n are given, to find a. =C; A= L(-1)+M and cor. if MPB, and mp ; See question 540. n = See Index. Cor. 1. When p, n, and a are given, to find m, first find r by 545, and then m by question 545. Cor. 2. Where m, n, and a are given, to find p, first find r, by 545, and then Ρ as above. = 5461. Annuities in reversion by Comp. Interest. Let a =annuity, p the present worth thereof, t time of its duration, x = time before it commences, and r = the interest of $1 in 1 year. Case 1. When a, t, x, are r, given to find p. then, as before, the present worth of a to xt or x years to {(1+r)2+1}a {(1+r)-1a; then 1}a, {(1+r)+' } a—a r(1+r)*+i {(1+r)*x}a- a r(1+r)* ift XL(1+r)=B; then A+L(b—1) — R—(x+t) × L(1+r) -P. When p, t, x and r are given, to find a, then AP+R +(x+t) × Ĺ(1+r) — L.(b—1). And also when p, a, t and r A+L(6-1) —R-P-B are given, to find x; then when p, a, x and r are given, to find t; if P+R+≈ × L(1+r)' Cor. 1. When p, a x and t are given to 12h find r. S Index. Let (2x+n+1)=h; u= t2--1 A+T-P h V; zu -2(v-1); then (Z-U)=E, and u e=r. 547. The amount of an annuity in arrear, computed at simple interest, was $215; now if the annuity had continued unpaid one year longer, it would have amounted to $275, but if one year less, to no more than $157 50 cts. The annuity, time, and rate. annuity; n the time; and r the interest of $1; b, $275 =c, and $157 50=d; then, by the ques(n-1)(n-2) tion, I have d= (n-1)a+ 1.2 Put a= also $215: = ・ar; b=na+: n(n-1) 1.2 -ar; |