41 In how many years would $6000 amount to $7470, allowing the rate of interest to be 7 per cent? 42. What is the Formula for finding the time in which the sum s would amount to the sum a, if the interest be at r per cent? 43. What principal would produce as much interest in 3 years, as $500 would in 4 years, the rate of interest in both cases being 6 per cent? Let x represent the principal; then represents the interest for one year; and 6x 100 120 dollars is the interest on $500 for 4 years. Then the Equation will be 44. At what rate per cent of interest would $525 produce the same amount of interest in 5 years, that $700 would produce at 5 per cent in 3 years? Let x represent the rate per cent; then 525x 100 2625.x 100 represents the interest on $525 for one year; and is the interest on $525 for 5 years; and 105 dollars is the interest on $700 for 3 years at 5 We shall then have the Equation This Equation will give a 4 per cent. 45. A person who possessed a capital of $70000, put the greater part of it at interest at 5 per cent, and the other part at 4 per cent. The interest on the whole was $3250 per annum. Required the two parts. Let a represent the greater part, and y the less; then we shall have for the first Equation 5x x+y=70000. Now, represents the interest on the greater part for one year, 100 at 5 per cent; and 4y 100 per cent. represents the interest on the less part for one year, at 4 From the two Equations we shall find = $45000, and y=$25000. 46. The sum of $200 is to be applied in part towards the payment of a debt of $300, and in part to paying the interest at 6 per cent in advance, for 12 months, on the remainder of the debt. What is the amount of the payment that can be made on the debt? Let x represent the payment; then 300--x represents the remainder of the debt; and 6 (300-x) represents the interest on the remainder. 100 47. A is indebted to B $1000, and is able to raise but $600. With this sum A proposes to pay part of the debt, and the interest, at 8 per cent in advance, on his note at 2 years for the remainder. For what sum should the note be drawn? Let a represent the sum; then 8x 100 16x 100 represents the interest on the note for one year; represents the interest on the note for 2 years; and 1000-x represents the present payment. From this Equation x will be found equal to $476.19'. 48. Find the Formulas for dividing the sum s into two parts, one of which is to be applied towards the payment of a debt of n dollars, and the other to paying the interest, in advance, on the remainder of the debt, for t years, at r per cent per annum? Let a represent the present payment; then n-x represents the remainder of the debt; and represents the interest on the remainder for t years, 1. The first term of an increasing Arithmetical Progression is 3, the common difference of the terms is 2, and the number of terms 20. What is the last term? and the sum of all the terms? The last term will be 3+2(20-1)=41, (176). The sum of all the terms will be (3+41)×20=440, (180). 2. The first term of a decreasing Arithmetical progression is 100, the common difference of the terms is 3, and the number of terms 34. What is the last term? and the sum of all the terms? The last term will be 100-3(34-1)=1, (176). The sum of all the terms will be (100+1)×34=1717, (180). 3. What is the sum of the numbers 1, 2, 3, 4, 5, &c., continued to 1000 terms? Here, the common difference is evidently 1; the number of terms 1000; and the last term 1000. Hence, the sum of all the terms will be (1+1000) × 1000-500500, (180). 4. What is the common difference of the terms in an Arithmetical progression whose first term is 10, last term 150, and number of terms 21? The common difference is (150-10)(21-1)=7, (177). 5. If the third term of an Arithmetical progression be 40, and the fifth term 70, what will the fourth term be? The fourth term will be 1(40+70)=55, (179). 6. If the first term of an Arithmetical progression be 5, and the fifth term 30, what will the second, third, and fourth terms be? The common difference of the terms is (30—5) ÷ (5—1)=6}, (177). Hence, the second term will be 5+61=111; the third 114+61= 17; and the fourth 171+61=233. 7. If the fourth term of an Arithmetical progression be 37, and the eighth term 60, what are the intermediate terms? The number of terms being 5, the common difference is (60-37)-(5-1)=53, (177) Hence, the fifth term is 37+53-423; the sixth is 423+53-48}; the seventh 481+52=541. 8. What is the sum of 25 terms of an increasing Arithmetical progression in which the first term is, and the common difference of the terms also ? The last term will be +(25-1)=121, (176). Hence, the sum of all the terms is (1+121)×25=1621, (180). 9. The first term of an increasing Arithmetical progression is 1, and the number of terms is 23. What must the common difference be, that the sum of all the terms may be 100? Let x represent the common difference; then 2 23 represents the sum of the terms, (180). We shall then have the Equation |