PROBLEM VI. To find the present value of an annuity a payable for t years. simple interest being allowed at the rate r. It is manifest that the present value of the annuity must be a sum such that, if put out at interest for t years at the rate r, its amount at the end of that period will be the same with the amount of the annuity. Hence, if we call this present value p, we shall have, by Problems I. and V., p+ptr amount of annuity. t(t-1) =ta+ra. 1.2 PROBLEM VII. To find the amount of a sum p laid out for t years, compound interest being allowed at the rate r. At the end of the first year the amount will be, by Problem II., p+pr, or p(1+r). Since compound interest is allowed, this sum p(1+r) now becomes the principal, and hence, at the end of the second year, the amount will be p(1+r), together with the interest on p(1+r) for one year; that is, it will be p(1+r)+pr(1+r), or p(1+r)2. The sum p(1+r) must now be considered as the principal, and hence the whole amount, at the end of the third year, will be p(1+r)2+pr(1+r)2, or p(1+r)3. And, in like manner, at the end of the tth year, we shall have Any three of the four quantities, s, p, r, t, being given, the fourth may always be found from the above equation. EXAMPLE I. Find the amount of $15.50 for 9 years, compound interest being allowed at the rate of 3 per cent. per annum, the interest payable at the end of each year. * It is unnecessary to give any examples under this rule, as the purchase of annuities at simple interest can never be of practical utility. EXAMPLE II. Find the amount of £182 12s. 6d. for 18 years, 6 months, and 10 days, at the rate of 3 per cent. per annum, compound interest, the interest being payable at the end of each year. In this case, it will be convenient, first, to find the amount at compound interest of the above sum for 18 years, and then calculate the interest on the result for the remaining period. By formula (7), Here s=p(1+r) log. s log. p+t log. (1+r) p=£182. 12s. 6d.=£182.625 T= t= ... =£.035 =18 years .log. p=2.2615602 t log. (1+r)=0.2689254 ... .log.s=2.5304856= log. of 339.224. Again, to find the interest on this sum for the short period, we have Required the compound interest upon $410 for 21 years at 4 per cent. per annum, the interest being payable half yearly. In this case the time t must be calculated in half years; and, since we have supposed r to be the interest of $1 for one year, we must substitute which will be the interest of $1 for half a year; the formula (7) will thus become 2' The interest must be the difference between this amount and the original principal; ..i-s-p = $458.247-$410 EXAMPLE IV. $400 was put out at compound interest, and at the end of 9 years amounted to $569.333; required the rate of interest per cent. In what time will a sum of money double itself, allowing 4 per cent. compound interest? In like manner, if it be required to find in what time a sum will triple itself at the same rate, we have PRESENT VALUE AND DISCOUNT AT COMPOUND INTEREST. If we call p the present value of a sum s due t years hence, and d its discount, reasoning precisely in the same manner as in the case of simple interest, we shall find PROBLEM VIII.-To find the amount of an annuity a continued for t years, compound interest being allowed at the rate r. At the end of the first year the annuity a will become due; at the end of the second year a second payment a will become due, together with the interest of the first payment a for one year, that is, ar; the whole sum upon which interest must now be computed is thus, 2a+ar. At the end of the third year a further payment a becomes due, together with the interest on 2a+ar, i. e., 2ar+ar2; the whole sum upon which interest must now be computed is 3a +3ar+ar2. The result will appear evident when exhibited under the following form: Whole amount at the end of second year, a+a+ar =a+a(1+r). Whole amount at the end of third year, a+a+a(1+r)+ar+ar(1+r) =a+a(1+r)+a(1+r)2. Whole amount at the end of fourth year, =a+a+a(1+r)+ a(1+r)2 + ar &c. &c. Whole amount at the end of th year, +ar(1+r)+ar(1+r)2. =a+a(1+r)+a(1+r)2+a(1+r)3 &c. =a+a(1+r)+a(1+r)2+a(1+r)3 +............a(1+r)'−1. Hence the whole amount is, in terms of the sum of a geometric progression, s=a{1+(1+r) + (1+r)2+...... •+(1+r) } PROBLEM IX.-To find the present value of an annuity a payable for t years, compound interest being allowed at the rate r. It is manifest that the present value of this annuity must be a sum such, that if put out at interest for t years at the rate r, its amount at the end of that period will be the same as the amount of the annuity. Hence, if we call this present value p, we shall have, by Probs. VII. and VIII., = (11) EXAMPLE. What is the present value of an annuity of $500, to last for 40 years, compound interest being allowed at the rate of 21 per cent. per annum. PROBLEM X.-To find the present value (P) of an annuity a which is to commence after T years, and to continue for t years. The present value required is manifestly the present value of a for T+! years, minus the present value of a for T years. a (1+r)+1 By Problem IX., the present value of a for T+t years =. By Problem IX., the present value of a for T years r (1+r)TM+t a (1+r)-1 = r (1+r)TM PROBLEM XI.-To find the present value p of an estate, or perpetuity, whose annual rental is a, compound interest being calculated at the rate r. The present value of an annuity a, to continue for t years, by Prob. IX., is P= 1 but if the annuity last forever, as in the case of an estate, then t=∞, and |