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41 In how many years would $6000 amount to $7470, allowing the rate of interest to be 7 per cent?

Let x represent the number of years; then

42000

6000 X

100

42000x

represents the interest for a years.

100

We shall then have the following Equation:

42000.x

6000+

=7470;

100

from which we shall find x=3 years.

7

100'

or

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?

Let x represent the number of years; then

ST

100

STX

100

100

42x

is the interest for one year; and

"

represents the interest for one year; and

The Equation will then be

represents the interest for x years.

s+

100(a-s)

sr

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which will give x=

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

6x

represents the interest for one year; and

represents the interest for 3 years; and

200

120 dollars is the interest on $500 for 4 years. Then the Equation will be

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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

represents the interest on $525 for one year; and

is the interest on $525 for 5

100

and

years;

105 dollars is the interest on $700 for 3 years at 5 We shall then have the Equation

per

2625.x

=105.

100

This Equation will give x= =4 per cent.

per cent.

Let a represent the greater part, and for the first Equation

x+y=70000.

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.

cent.

5x

Now, represents the interest on the greater part for one year,

100

at 5 per cent; and

4y

100

The second Equation will then be

5x

100

y the less; then we shall have

represents the interest on the less part for one year, at 4

4y

+

=3250.

100

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

We shall then have the Equation

x+ (300-x)

from which we shall find a=$193.61'.

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 x represent the sum; then

8x

100

16x

represents the interest on the note for one year;

Therefore, the Equation is

represents the interest on the note for 2 years; and 1000 - represents the present payment..

100

6

100

at r per cent.

= 200;

16.x

(1000-x)+ =600.

100

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 x represent the present payment; then

n-x represents the remainder of the debt; and

rt

(n-x)

100

Hence, the Equation will be

x+(n-x)

represents the interest on the remainder for t years,

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The sum of all the terms will be

Problems in Progressions.

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).

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100s-nrt
100-rt

the present pay

the interest on the remainder.

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).

(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).

The common difference is

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?

(150-10)(21-1)=7, (177).

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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=114; the third 11+6= 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+53-54.

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=162, (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 a represent the common difference; then

1+22x represents the last term (176); and
2+22x

×23 represents the sum of the terms, (180).
We shall then have the Equation

2

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