This will give a quadratic equation in x, namely, P ̧rx3 — {P ̧r (t ̧ + t2) + P2+ P ̧} x + P ̧rt ̧t2+ P ̧t2+P ̧t ̧=0; 1 12 11 that root must be taken which lies between t, and t 22 581. Another method of solving the question of the preceding article is as follows: The present value of P, due at the end of t1 years is 1 the present value of P, due at the end of t, years is 2 P, 1 1 + tri the present value of P1+ P, due at the end of x years is 1 1 2 1 + xr Hence we may propose to find the equated time of payment, x, from the equation 582. If such a question did occur in practice however the method would probably be to proceed as in the first solution, with this exception, that the lender would allow interest instead of discount on the sum paid before it was due; thus we should find x from 1 1 2 In this case the interest on P1 + P2 for x years is equal to the sum of the interests of P, and P, for the times t1 and t ̧ respectively; this follows if we multiply both sides of the last equation by r. This rule is more advantageous to the borrower than that in Art. 580, for the interest on a given amount is greater than the discount. See Art. 577. due at 583. Suppose there are several sums P,, P P the end of times t,, t,, t2, ...... respectively, and the equated time of payment is required. The first method of solution (Art. 580) becomes very complicated in this case, and we shall therefore omit it. The second method (Art. 581) gives for determining the equated time x, 584. Equation of payments is a subject of no practical importance, and seems retained in books chiefly on account of the apparent paradox of different methods occurring which may appear equally fair, but which lead to different results. We refer the student for more information on the question to the article Discount in the English Cyclopædia. We may observe, however, that the difficulty, if such it be, arises from the fact that simple interest is almost a fiction; the moment any sum of money is due, it matters not whether it is called principal or interest, it is of equal value to the owner; and thus if the interest on borrowed money is retained by the borrower, it ought in justice to the lender, to be united to the principal, and charged with interest afterwards. 585. If compound interest be allowed, the solutions in Arts. 580 and 581 will give the same result. For the solution according to Art. 580 will be as follows: therefore 2 the interest on P, for x-t, years is P1 (R*—tı — 1); From this equation x must be found; by transposition we shall see that this is the same equation as would be obtained by the method of Art. 581; for we obtain which shews that x is such that the present value of P1+P ̧ due at the end of x years is equal to the sum of the present values of P, and P, due at the end of t, and t ̧ years respectively. 1 2 ...... I due at the 586. If there be different sums P1, P., P. end of t1, t, t,..... years respectively, the equated time of payment, x, allowing compound interest, may be found from 587. We have said in Art. 580, that we must take that root of the quadratic equation which lies between t, and t1; we will now prove that there will in fact be always one root, and only one, between t, and t We have to shew that the equation has one root, and only one, lying between t, and t The expression P ̧(x − t ̧) {1 + (t ̧ − x) r} − P ̧ (t ̧ − x) 2 is obviously positive when x=t. If this expression is arranged in the form ax2 + bx + c; the coefficient a is negative, being - P1r; hence t, must lie between the roots of the equation by Art. 339; that is, one root is greater than t, and one root less than t It is obvious too that no value of x less than t, can make the expression vanish, so there cannot be a root of the equation less than t1; there must then be one root between t, and t,, and one root greater than t 1 It may be remarked that the value x=t ̧+ also makes the expression positive, and so the root which is greater than t, must 1. Find the equated time of payment of two sums, one of £400 due two years hence, the other of £2100 due eight years hence, at 5 per cent. (Art. 580.) 2. Find the equated time of payment of two sums, one of £20 due at the present date, the other of £16. 5s. due 270 days hence, the rate of interest being twopence-halfpenny per hundred pounds per day. (Art. 580.) 3. Find the equated time of paying two sums of money due at different epochs, interest being supposed due every moment. 4. A sum of money is left by will to be divided into three parts such that their amounts at compound interest, in a, b, c years respectively, shall be equal; determine the parts. 5. If a and n be positive integers, the integral part of {a + √(a2-1)}" is odd. 6. If a and n be positive integers, the integral part of {√(a2 + 1) + a}" is odd when n is even, and even when n is odd. T. A. 23 7. Shew that the remainder after n terms of the expansion of 2 α (2) 8. If y (n,r) = n (n − 1) (n − 2) ... (n−r + 1), shew that ¥ (n, r) = 4 (n − 2, r) + 2r¥ (n − 2, r − 1) + r (r − 1) 4 (n − 2, r − 2). 9. If ¿ (n, r) – n (n − 1). ... (n − r + 1), shew that = L $(n, m) = 4 (n − m + 1, 1) + p (m − 1, 1) † (n − m + 1, 2) a +4(m-1, 2) + (n − m + 1, 3) +.. 10. With the same notation shew that − (a + ß) & (n, 1) + (a + 2ß) + (n, 2) − (a + 3ß) † (n, 3) + 11. If s be the sum of n terms of a geometric progression whose first term is a and common ratio 1+x, where x is very small, shew that 12. If a quantity change continuously in value from a to b in a given time t1, the increase at any instant bearing a constant ratio to its value at that instant, prove that its value at any time t will be a t ()". (Art. 574.) XLIII. ANNUITIES. 588. To find the amount of an annuity left unpaid for any number of years, allowing simple interest upon each sum from the time it becomes due. Let A be the annuity, n the number of years, r the interest of one pound for one year, M the amount. |