Math Problems And Solutions 4n4u2p

Math Problems and Solutions on Integers Problems related to integer numbers are presented along with their solutions. Problem 1: Find two consecutive integers whose sum is equal 129. Solution to Problem 1: 

Let x and x + 1 (consecutive integers differ by 1) be the two numbers. Use the fact that their sum is equal to 129 to write the equation x + (x + 1) = 129

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Solve for x to obtain x = 64

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The two numbers are x = 64 and x + 1 = 65

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We can see that the sum of the two numbers is 129.

Problem 2: Find three consecutive integers whose sum is equal to 366. Solution to Problem 2: 

Let the three numbers be x, x + 1 and x + 2. their sum is equal to 366, hence x + (x + 1) + (x + 2) = 366



Solve for x and find the three numbers x = 121 , x + 1 = 122 and x + 2 = 123

Problem 3: The sum of three consecutive even integers is equal to 84. Find the numbers. Solution to Problem 3: 

The difference between two even integers is equal to 2. let x, x + 2 and x + 4 be the three numbers. Their sum is equal to 84, hence x + (x + 2) + (x + 4) = 84

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Solve for x and find the three numbers x = 26 , x + 2 = 28 and x + 4 = 30

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The three numbers are even. Check that their sum is equal to 84.

Problem 4: The sum on an odd integer and twice its consecutive is equal to equal to 3757. Find the number. Solution to Problem 4: 

The difference between two odd integers is equal to 2. let x be an odd integer and x + 2 be its consecutive. The sum of x and twice its consecutive is equal to 3757 gives an equation of the

form x + 2(x + 2) = 3757 

Solve for x x = 1251



Check that the sum of 1251 and 2(1251 + 2) is equal to 3757.

Problem 5: The sum of the first and third of three consecutive even integers is 131 less than three times the second integer. Find the three integers. Solution to Problem 5: 

Let x, x + 2 and x + 4 be three integers. The sum of the first x and third x + 4 is given by x + (x + 4)

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131 less than three times the second 3(x + 2) is given by 3(x + 2) - 131

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"The sum of the first and third is 131 less than three times the second" gives x + (x + 4) = 3(x + 2) - 131

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Solve for x and find all three numbers x = 129 , x + 2 = 131 , x + 4 = 133

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As an exercise, check that the sum of the first and third is 131 less than three times

Problem 6: The product of two consecutive odd integers is equal to 675. Find the two integers. 

Let x, x + 2 be the two integers. Their product is equak to 144 x (x + 2) = 675



Expand to obtain a quadratic equation. 2

x + 2 x - 675 = 0 

Solve for x to obtain two solutions x = 25 or x = -27 if x = 25 then x + 2 = 27 if x = -27 then x + 2 = -25



We have two solutions. The two numbers are either 25 and 27



or -27 and -25

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Check that in both cases the product is equal to 675.

Problem 7: Find four consecutive even integers so that the sum of the first two added to twice the sum of the last two is equal to 742. Solution to Problem 7: 

Let x, x + 2, x + 4 and x + 6 be the four integers. The sum of the first two x + (x + 2)



twice the sum of the last two is written as 2 ((x + 4) + (x + 6)) = 4 x + 20

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sum of the first two added to twice the sum of the last two is equal to 742 is written as x + (x + 2) + 4 x + 20 = 742

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Solve for x and find all four numbers x = 120 , x + 2 = 122 , x + 4 = 124 , x + 6 = 126

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As an exrcise, check that the sum of the first two added to twice the sum of the last two is equal to 742

Problem 8: When the smallest of three consecutive odd integers is added to four times the largest, it produces a result 729 more than four times the middle integer. Find the numbers and check your answer. Solution to Problem 8: 

Let x, x + 2 and x + 4 be the three integers. "The smallest added to four times the largest is written as follows" x + 4 (x + 4)

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"729 more than four times the middle integer" is written as follows 729 + 4 (x + 2)

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"When the smallest is added to four times the largest, it produces a result 729 more than four times the middle" is written as follows x + 4 (x + 4) = 729 + 4 (x + 2)

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Solve for x and find all three numbers x + 4 x + 16 = 729 + 4 x + 8 x = 721 x + 2 = 723 x + 4 = 725

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Check: the smallest is added to four times the largest 721 + 4 * 725 = 3621

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four times the middle 4 * 723 = 2892

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3621 is more than 2892 by 3621 - 2892 = 729

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The answer to the problem is correct.

Mixture Problems With Solutions Mixture problems and their solutions are presented along with their solutions. Percentages are also used to solve these types of problems. Problem 1: How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution? Solution to Problem 1: 

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hence x + 40 = y



We shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But the alcohol is measured in percentage term. 20% x + 50% * 40 = 30% y



Substitute y by x + 40 in the last equation to obtain. 20% x + 50% * 40 = 30% (x + 40)

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Change percentages into fractions. 20 x / 100 + 50 * 40 / 100= 30 x / 100 + 30 * 40 / 100

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Mutliply all by 100 to simplify. 20 x + 50 * 40 = 30 x + 30 * 40

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Solve for x. x = 80 liters



80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution.

Problem 2: John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a 2% alcohol solution with a 7% alcohol solution. What are the quantities of each of the two solutions (2% and 7%) he has to use?

Solution to Problem 2: 

Let x and y be the quatities of the 2% and 7% aclohol solutions to be used to make 100 ml. Hence x + y = 100



We now write mathematically that the quantity of alcohol in x ml plus the quantity of alcohol in y ml is equal to the quantity of alcohol in 100 ml. 2% x + 7% y = 5% 100

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The first equation gives y = 100 - x. Substitute in the last equation to obtain 2% x + 7% (100 - x) = 5% 100

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Multiply by 100 and simplify 2 x + 700 - 7 x = 5 * 100

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Solve for x x = 40 ml

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Substitute x by 40 in the first equation to find y y = 100 - x = 60 ml

Problem 3: Sterling Silver is 92.5% pure silver. How many grams of Sterling Silver must be mixed to a 90% Silver alloy to obtain a 500g of a 91% Silver alloy? Solution to Problem 3: 

Let x and y be the weights, in grams, of sterling silver and of the 90% alloy to make the 500 grams at 91%. Hence x + y =500

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The number of grams of pure silver in x plus the number of grams of pure silver in y is equal to the number of grams of pure silver in the 500 grams. The pure silver is given in percentage forms. Hence 92.5% x + 90% y = 91% 500

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Substitute y by 500 - x in the last equation to write 92.5% x + 90% (500 - x) = 91% 500

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Simplify and solve 92.5 x + 45000 - 90 x = 45500 x = 200 grams.

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200 grams of Sterling Silver is needed to make the 91% alloy.

Problem 4: How many Kilograms of Pure water is to be added to 100 Kilograms of a 30% saline solution to make it a 10% saline solution.

Solution to Problem 4: 

Let x be the weights, in Kilograms, of pure water to be added. Let y be the weight, in Kilograms, of the 10% solution. Hence x + 100 = y



Let us now express the fact that the amount of salt in the pure water (which 0) plus the amount of salt in the 30% solution is equal to the amount of salt in the final saline solution at 10%. 0 + 30% 100 = 10% y

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Substitute y by x + 100 in the last equation and solve. 30% 100 = 10% (x + 100)

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Solve for x. x = 200 Kilograms.

Problem 5: A 50 ml after-shave lotion at 30% alcohol is mixed with 30 ml of pure water. What is the percentage of alcohol in the new solution? Solution to Problem 5: 

The amount of the final mixture is given by 50 ml + 30 ml = 80 ml

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The amount of alcohol is equal to the amount of alcohol in pure water ( which is 0) plus the amount of alcohol in the 30% solution. Let x be the percentage of alcohol in the final solution. Hence 0 + 30% 50 ml = x (80)

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Solve for x x = 0.1817 = 18.75%

Problem 6: You add x ml of a 25% alcohol solution to a 200 ml of a 10% alcohol solution to obtain another solution. Find the amount of alcohol in the final solution in of x. Find the ratio, in of x, of the alcohol in the final solution to the total amount of the solution. What do you think will happen if x is very large? Find x so that the final solution has a percentage of 15%. Solution to Problem 6: 

Let us first find the amount of alcohol in the 10% solution of 200 ml. 200 * 10% = 20 ml

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The amount of alcohol in the x ml of 25% solution is given by 25% x = 0.25 x

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The total amount of alcohol in the final solution is given by 20 + 0.25 x

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The ratio of alcohol in the final solution to the total amount of the solution is given by [ ( 20 + 0.25 x ) / (x + 200)]

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If x becomes very large in the above formula for the ratio, then the ratio becomes close to 0.25 or 25% (The above function is a rational function and 0.25 is its horizontal asymptote). This means that if you increase the amount x of the 25% solution, this will dominate and the final solution will be very close to a 25% solution.

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To have a percentage of 15%, we need to have [ ( 20 + 0.25 x ) / (x + 200)] = 15% = 0.15

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Solve the above equation for x 20 + 0.25 x = 0.15 * (x + 200) x = 300 ml

Rate, Time Distance Problems With Solutions The relationship between distance, rate (speed) and time Distance = time * rate is used to solve uniform motion problems. Detailed solutions to the problems are provided. Problem 1: Two cars started from the same point, at 5 am, traveling in opposite directions at 40 and 50 mph respectively. At what time will they be 450 miles apart? Solution to Problem 1: 

After t hours the distances D1 and D2, in miles per hour, traveled by the two cars are given by D1 = 40 t and D2 = 50 t

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After t hours the distance D separating the two cars is given by D = D1 + D2 = 40 t + 50 t = 90 t

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Distance D will be equal to 450 miles when D = 90 t = 450 miles

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To find the time t for D to be 450 miles, solve the above equation for t to obtain t = 5 hours. 5 am + 5 hours = 10 am

Problem 2: At 9 am a car (A) began a journey from a point, traveling at 40 mph. At 10 am another car (B) started traveling from the same point at 60 mph in the same direction as car (A). At what time will car B car A? Solution to Problem 2:



After t hours the distances D1 traveled by car A is given by D1 = 40 t

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Car B starts at 10 am and will therefore have spent one hour less than car A when it es it. After (t - 1) hours, distance D2 traveled by car B is given by D2 = 60 (t-1)

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When car B es car A, they are at the same distance from the starting point and therefore D1 = D2 which gives 40 t = 60 (t-1)

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Solve the above equation for t to find t = 3 hours

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Car B es car A at 9 + 3 = 12 pm

Problem 3: Two trains, traveling towards each other, left from two stations that are 900 miles apart, at 4 pm. If the rate of the first train is 72 mph and the rate of the second train is 78 mph, at wthat time will they each other? Solution to Problem 3: 

After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t

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After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t + 78 t = 150 t

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When distance D is equal to 900 miles, the two trains each other. 150 t = 900

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Solve the above equation for t t = 6 hours.

Problem 4: John left home and drove at the rate of 45 mph for 2 hours. He stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination. How many miles did John drive? Solution to Problem 4: 

The total distance D traveled by John is given by D = 45 * 2 + 3 * 55 = 255 miles.

Problem 5: Linda left home and drove for 2 hours. She stopped for lunch then drove for another 3 hours at a rate that is 10 mph higher than the rate before she had lunch. If the total distance Linda travelled is 230 miles, what was the rate before lunch?

Solution to Problem 5: 

If x is the rate at which Linda drove before lunch the rate after lunch is equal x + 10. The total distance D traveled by Linda is given by D = 2 x + 3(x + 10)



and is equal to 230 miles. Hence 2 x + 3 (x + 10) = 230

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Solve for x to obtain x = 40 miles / hour.

Problem 6: Two cars left, at 8 am, from the same point, one traveling east at 50 mph and the other travelling south at 60 mph. At what time will they be 300 miles apart? Solution to Problem 6: 

A diagram is shown below to help you understand the problem.

.



The two cars are traveling in directions that are at right angle. Let x and y be the distances traveled by the two cars in t hours. Hence x = 50 t and y = 60 t

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Since the two directions are at right angle, Pythagora's theorem can used to find distance D between the two cars as follows: 2

2

D = sqrt ( x + y ) 

We now find the time at which D = 300 miles by solving 2

2

sqrt ( x + y ) = 300



Square both sides and substitute x and y by 50 t and 60 t respectively to obtain the equation 2

2

(50 t) + (60 t) = 300 

2

Solve the above equations to obtain t = 3.84 hours (rounded to two decimal places) or 3 hours and 51 minutes (to the nearest minute)



The two cars will 300 miles apart at 8 + 3 h 51' = 11:51 am.

Problem 7: By Car, John traveled from city A to city B in 3 hours. At a rate that was 20 mph higher than John's, Peter traveled the same distance in 2 hours. Find the distance bewteen the two cities. Solution to Problem 7: 

Let x be John's rate in traveling between the two cities. The rate of Peter will be x + 10. We use the rate-time-distance formula to write the distance D traveled by John and Peter (same distance D) D = 3 x and D = 2(x + 20)

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The first equation can be solved for x to give x=D/3

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Substitute x by D / 3 into the second equation D = 2(D/3 +20)

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Solve for D to obtain D = 120 miles

Problem 8: Gary started driving at 9:00 am from city A towards city B at a rate of 50 mph. At a rate that is 15 mph higher than Gary's, Thomas started driving at the same time as John from city B towards city A. If Gary and Thomas crossed each other at 11 am, what is the distance beween the two cities? Solution to Problem 8: 

Let D be the distance between the two cities. When Gary and Thomas cross each other, they have covered all the distance between the two cities. Hence D1 = 2 * 50 = 100 miles , distance traveled by Gary D1 = 2 * (50 + 15) = 130 miles , distance traveled by Gary

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Distance D between the two cities is given by D = 100 miles + 130 miles = 230 miles

Problem 9: Two cars started at the same time, from the same point, driving along the same road. The rate of the first car is 50 mph and the rate of the second car is 60 mph. How long will it take for the distance bewteen the two cars to be 30 miles? Solution to Problem 9:



Let D1 and D2 be the distances traveled by the two cars in t hours D1 = 50 t and D2 = 60 t

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The second has a higher speed and therefore the distance d between the two cars is given by d = 60 t - 50 t = 10 t

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For d to be 30 miles, we need to have 30 miles = 10 t

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Solve the above equation for t to obtain t = 3 hours.

Problem 10: Two trains started at 10 pm, from the same point. The first train traveled North at the rate of 80 mph and the second train traveled South at the rate of 100 mph. At what time were they 450 miles apart? Solution to Problem 10: 

Let D1 and D2 be the distances traveled by the two trains in t hours. D1 = 80 t and D2 = 100 t

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Since the two trains are traveling in opposite directions, then total distance D between the two trains is given by D = D1 + D2 = 180 t

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For this distance to be 450 miles, we need to have 180 t = 450

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Solve for t to obtain t = 2 hours 30 minutes. 10 pm + 2:30 = 12:30 am

Problem 11: Two trains started from the same point. At 8:00 am the first train traveled East at the rate of 80 mph. At 9:00 am, the second train traveled West at the rate of 100 mph. At what time were they 530 miles apart? Solution to Problem 11: 

When the first train has traveled for t hours the second train will have traveled (t - 1) hours since it strated 1 hour late. Hence if D1 and D2 are the distances traveled by the two trains, then D1 = 80 t and D2 = 100 (t - 1)

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Since the trains are traveling in opposite directios, the total distance D between the two trains is given by D = D1 + D2 = 180 t - 100

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For D to be 530 miles, we need to have 180 t - 100 = 530

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Solve for t t = 3 hours 30 minutes. 8 am + 3:30 = 11:30 am Work Rate Problems with Solutions

A set of problems related to work and rate of work is presented with detailed solutions. Problem 1: It takes 1.5 hours for Tim to mow the lawn. Linda can mow the same lawn in 2 hours. How long will it take John and Linda, work together, to mow the lawn? Solution to Problem 1: 

We first calculate the rate of work of Jonh and Linda John: 1 / 1.5 and Linda 1 / 2

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Let t be the time for John and Linda to mow the Lawn. The work done by John alone is given by t * (1 / 1.5)

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The work done by Linda alone is given by t * (1 / 2)

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When the two work together, their work will be added. Hence t * (1 / 1.5) + t * (1 / 2) = 1

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Multiply all by 6 6 (t * (1 / 1.5) + t * (1 / 2) ) = 6

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and simplify 4t+3t=6

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Solve for t t = 6 / 7 hours = 51.5 minutes.

Problem 2: It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B used alone takes 8 hours to fill the same tank. We want to use three pumps: A, B and another pump C to fill the tank in 2 hours. What should be the rate of pump C? How long would it take pump C, used alone, to fill the tank? Solution to Problem 2: 

The rates of pumps A and B can be calculated as follows: A: 1 / 6 and B: 1 / 8

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Let R be the rate of pump C. When working together for 2 hours, we have 2(1/6+1/8+R)=1

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Solve for R R = 1 / 4.8 , rate of pump C.

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Let t be the time it takes pump C, used alone, to fill the tank. Hence t * (1 / 4.8) = 1

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Solve for t t = 4.8 hours , the time it takes pump C to fill the tank.

Problem 3: A tank can be filled by pipe A in 5 hours and by pipe B in 8 hours, each pump working on its own. When the tank is full and a drainage hole is open, the water is drained in 20 hours. If initially the tank was empty and somenone started the two pumps together but left the drainage hole open, how long does it take for the tank to be filled? Solution to Problem 3: 

Let's first find the rates of the pumps and the drainage hole pump A: 1 / 5 , pump B: 1 / 8 , drainage hole: 1 / 20



Let t be the time for the pumps to fill the tank. The pumps ,add water into the tank however the drainage hole drains water out of the tank, hence t ( 1 / 5 + 1 / 8 - 1 / 20) = 1

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Solve for t t = 3.6 hours.

Problem 4: A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pump working on its own. At 9 am pump A is started. At what time will the swimming pool be filled if pump B is started at 10 am? Solution to Problem 4: 

the rates of the two pumps are pump A: 1 / 3 , pump B: 1 / 6



Working together, If pump A works for t hours then pump B works t - 1 hours since it started 1 hour late. Hence t * (1 / 3) + (t - 1) * (1 / 6) = 1

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Solve for t t = 7 / 3 hours = 2.3 hours = 2 hours 20 minutes.

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The swimming pool will be filled at 9 + 2:20 = 11:20 Projectile Problems with Solutions

A tutorial on solving projectile problems. Detailed solutions to the problems are provided. Review If air resistance is ignored, the height h of a projectile above the ground after t seconds is given by 2

H(t) = - (1 / 2) g t + Vo t + Ho 2

where g is the acceleration due to gravity which on earth is approximately equal to 32 feet / s , Vo is the initial velocity (when t = 0 ) and Ho is the initial height (when t = 0). If an object is dropped, the distance traveled by this object is given by H(t) = (1 / 2) g t

2

For a quadratic function of the form 2

H(t) = a t + b t + c

and with a negative, the maximum of H occur at t = - b / 2a (vertex of the graph of H).

Problem 1: 2

The formula h (t) = -16 t + 32 t + 80 gives the height h above ground, in feet, of an object thrown, at t = 0, straight upward from the top of an 80 feet building.

a - What is the highest point reached by the object? b - How long does it take the object to reach its highest point? c - After how many seconds does the object hit the ground? d - For how many seconds is the hight of the object higher than 90 feet? Solution to Problem 1: 

a - The height h given above is a quadratic function. The graph of h as a function of time t gives a parabolic shape and the maximum height h occur at the vertex of the parabola. For a 2 quadratic function of the form h = a t + b t + c, the vertex is located at t = - b / 2a. Hence for h given above the vertex is at t t = -32 / 2(-16) = 1 second.



1 second after the object was thrown, it reaches its highest point (maximum value of h) which is given by 2

h = -16 (1) + 32 (1) + 80 = 96 feet 

b - It takes 1 second for the object to reach it highest point.



c - At the ground h = 0, hence the solution of the equation h = 0 gives the time t at which the object hits the ground. 2

-16 t + 32 t + 80 = 0 

The above quadratic equation has two solutions one is negative and the second one is positive and approximately equal to 3.5 seconds. So it takes 3.5 seconds fopr the object to hit the ground after it has been thrown upward. The graphical meanings to the answers to parts a, b and c are shown below.

.



d - The object is higher than 90 feet for all values of t statisfying the inequality h > 90. Hence 2

-16 t + 32 t + 80 > 90 

The above inequality is satisfied for 0.4 < t < 1.6 (seconds)



The height of the object is higher than 90 feet for 1.6 - 0.4 = 1.2 seconds.



The graph below is that of h in of t and clearly shows that h is greater than 90 feet for t between 0.4 and 1.6 seconds.

. 2

Problem 2: A rock is dropped into a well and the distance travelled is 16 t feet, where t is the time. If the water splash is heard 3 seconds after the rock was dropped, and that the speed of sound is 1100 ft / sec, approximate the height of the well. Solution to Problem 2: 

Let T1 be the time it takes the rock to reach the bottom of the well. If H is the height of the well, we can write H = 16 T1



2

Let T2 be the time it takes sound wave to reach the top of the well. we can write H = 1100 T2



The relationship between T1 and T2 is T1 + T2 = 3



2

Eliminate H and combine the equations H = 16 T1 and H = 1100 T2 to obtain 2

16 T1 = 1100 T2 

We now substitute T2 by 3 - T1 in the above equation 2

16 T1 = 1100 (3 - T1) 

The above is a quadratic equation that may be written as 2

16 T1 + 1100 T1 - 3300 = 0 

The above equations has two solutions and only one of them is positive and is given by T1 = 2.88 seconds (2 decimal places)



We now calculate the height H of the well 2

2

H = 16 T1 = 16 (2.88) = 132.7 feet

Problem 3: From the top of a building, an object is thrown upward with an initial speed of 64 ft/sec. It touches the ground 5 seconds later. What is the height of the bulding Solution to Problem 3: 

The formula for the height H of a projectile thrown upward is given by 2

H(t) = -(1 / 2) g t + Vo t + Ho



g is a constant and equal to 32. The initial speed Vo is known and also when t = 5 seconds H = 0 (touches the ground). Ho is the initial height or height of the building. Hence we can write 2

0 = -(1 / 2) (32) (5) + (64) (5) + Ho 

Solve the above for Ho to find the height of the building Ho = 80 feet

Problem 4: From ground, an object is thrown upward with an initial speed Vo. Three seconds later, it reaches a maximum height. What is the initial velocity Vo? Solution to Problem 4: 

The height H of a projectile thrown upward from ground (initial height = 0 or Ho = 0) is given by 2

H(t) = -(1 / 2) (32) t + Vo t

2

H(t) = -16 t + Vo t



The maximum height occurs at t = -Vo / 2(-16). But this time is given and is equal to 3 seconds. Hence -Vo / 2(-16) = 3



Solve the above to find Vo = 96 feet / sec Mathematical Induction - Problems With Solutions

Several problems with detailed solutions on mathematical induction are presented. The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality…) is true for all positive integer numbers greater than or equal to some integer N. Let us denote the proposition in question by P (n), where n is a positive integer. The proof involves two steps: Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n. Step 2: We assume that P (k) is true and establish that P (k+1) is also true Problem 1: Use mathematical induction to prove that 1 + 2 + 3 + ... + n = n (n + 1) / 2 for all positive integers n. Solution to Problem 1: 

Let the statement P (n) be 1 + 2 + 3 + ... + n = n (n + 1) / 2



STEP 1: We first show that p (1) is true. Left Side = 1 Right Side = 1 (1 + 1) / 2 = 1



Both sides of the statement are equal hence p (1) is true.



STEP 2: We now assume that p (k) is true 1 + 2 + 3 + ... + k = k (k + 1) / 2



and show that p (k + 1) is true by adding k + 1 to both sides of the above statement

1 + 2 + 3 + ... + k + (k + 1) = k (k + 1) / 2 + (k + 1) = (k + 1)(k / 2 + 1) = (k + 1)(k + 2) / 2 

The last statement may be written as 1 + 2 + 3 + ... + k + (k + 1) = (k + 1)(k + 2) / 2



Which is the statement p(k + 1).

Problem 2: Prove that 2

2

2

2

1 + 2 + 3 + ... + n = n (n + 1) (2n + 1)/ 6 For all positive integers n. Solution to Problem 2: 

Statement P (n) is defined by 2

2

2

2

1 + 2 + 3 + ... + n = n (n + 1) (2n + 1)/ 2 

STEP 1: We first show that p (1) is true. 2

Left Side = 1 = 1 Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1 

Both sides of the statement are equal hence p (1) is true.



STEP 2: We now assume that p (k) is true 2

2

2

2

1 + 2 + 3 + ... + k = k (k + 1) (2k + 1)/ 6 

2

and show that p (k + 1) is true by adding (k + 1) to both sides of the above statement 2

2

2

2

2

1 + 2 + 3 + ... + k + (k + 1) = k (k + 1) (2k + 1)/ 6 + (k + 1) 

Set common denominator and factor k + 1 on the right side = (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6



Expand k (2k + 1)+ 6 (k + 1) 2

= (k + 1) [ 2k + 7k + 6 ] /6 

2

Now factor 2k + 7k + 6. = (k + 1) [ (k + 2) (2k + 3) ] /6

2



We have started from the statement P(k) and have shown that 2

2

2

2

2

1 + 2 + 3 + ... + k + (k + 1) = (k + 1) [ (k + 2) (2k + 3) ] /6 

Which is the statement P(k + 1).

Problem 3: Use mathematical induction to prove that 3

3

3

3

2

2

1 + 2 + 3 + ... + n = n (n + 1) / 4 for all positive integers n. Solution to Problem 3: 

Statement P (n) is defined by 3

3

3

3

2

2

1 + 2 + 3 + ... + n = n (n + 1) / 4 

STEP 1: We first show that p (1) is true. 3

Left Side = 1 = 1 2

2

Right Side = 1 (1 + 1) / 4 = 1 

hence p (1) is true.



STEP 2: We now assume that p (k) is true 3

3

3

3

2

2

1 + 2 + 3 + ... + k = k (k + 1) / 4 

3

add (k + 1) to both sides 3

3

3

3

3

2

2

1 + 2 + 3 + ... + k + (k + 1) = k (k + 1) / 4 + (k + 1) 

3

2

factor (k + 1) on the right side 2

2

= (k + 1) [ k / 4 + (k + 1) ] 

set to common denominator and group 2

2

= (k + 1) [ k + 4 k + 4 ] / 4 2

2

= (k + 1) [ (k + 2) ] / 4 

We have started from the statement P(k) and have shown that 3

3

3

3

3

2

2

1 + 2 + 3 + ... + k + (k + 1) = (k + 1) [ (k + 2) ] / 4 

Which is the statement P(k + 1).

Problem 4: 3

Prove that for any positive integer number n , n + 2 n is divisible by 3 Solution to Problem 4: 

Statement P (n) is defined by 3

n + 2 n is divisible by 3 

3

STEP 1: We first show that p (1) is true. Let n = 1 and calculate n + 2n 3

1 + 2(1) = 3 3 is divisible by 3 

hence p (1) is true.



STEP 2: We now assume that p (k) is true 3

k + 2 k is divisible by 3 is equivalent to 3

k + 2 k = 3 M , where M is a positive integer. 

3

We now consider the algebraic expression (k + 1) + 2 (k + 1); expand it and group like 3

3

2

(k + 1) + 2 (k + 1) = k + 3 k + 5 k + 3 3

2

= [ k + 2 k] + [3 k + 3 k + 3] 2

2

=3M+3[k +k+1]=3[M+k +k+1] 

3

Hence (k + 1) + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.

Problem 5: n

2

n

2

Prove that 3 > n for n = 1, n = 2 and use the mathematical induction to prove that 3 > n for n a positive integer greater than 2. Solution to Problem 5: 

Statement P (n) is defined by n

3 >n 

2

1

1

3 =3 2

1 =1 

2

STEP 1: We first show that p (1) is true. Let n = 1 and calculate 3 and 1 and compare them

3 is greater than 1 and hence p (1) is true.



Let us also show that P(2) is true. 2

3 =9 2

2 =4 

Hence P(2) is also true.



STEP 2: We now assume that p (k) is true k

3 >k 

2

Multiply both sides of the above inequality by 3 k

3*3 >3*k 

2

The left side is equal to 3 2

k+1

. For k >, 2, we can write

2

k > 2 k and k > 1 

We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities 2

2k >2k+1 

2

We now add k to both sides of the above inequality to obtain the inequality 2

2

3k >k +2k+1 

Factor the right side we can write 2



3 * k > (k + 1)

2

k

2

3 * 3 > (k + 1)

2

k



Rewrite the left side as 3 3



2

If 3 * 3 > 3 * k and 3 * k > (k + 1)

k+1

> (k + 1)

2 then

k+1

2

Which proves tha P(k + 1) is true

Problem 6: n

Prove that n ! > 2 for n a positive integer greater than or equal to 4. (Note: n! is n factorial and is given by 1 * 2 * ...* (n-1)*n.) Solution to Problem 6: 

Statement P (n) is defined by n! > 2



n

n

STEP 1: We first show that p (4) is true. Let n = 4 and calculate 4 ! and 2 and compare them

4! = 24 4

2 = 16 

24 is greater than 16 and hence p (4) is true.



STEP 2: We now assume that p (k) is true k! > 2



k

Multiply both sides of the above inequality by k + 1 k

k! (k + 1)> 2 (k + 1) 

The left side is equal to (k + 1)!. For k >, 4, we can write k+1>2



k

Multiply both sides of the above inequality by 2 to obtain k

2 (k + 1) > 2 * 2 

The above inequality may be written k

2 (k + 1) > 2 

k+1

k

k

We have proved that (k + 1)! > 2 (k + 1) and 2 (k + 1) > 2 (k + 1)! > 2



k

k+1

we can now write

k+1

We have assumed that statement P(k) is true and proved that statment P(k+1) is also true.

Problem 7: Use mathematical induction to prove De Moivre's theorem n

n

[ R (cos t + i sin t) ] = R (cos nt + i sin nt)

for n a positive integer. Solution to Problem 7: 

STEP 1: For n = 1 1

1

[ R (cos t + i sin t) ] = R (cos 1*t + i sin 1*t) 

It can easily be seen that the two sides are equal.



STEP 2: We now assume that the theorem is true for n = k, hence k

k

[ R (cos t + i sin t) ] = R (cos kt + i sin kt) 

Multiply both sides of the above equation by R (cos t + i sin t) k

k

[ R (cos t + i sin t) ] R (cos t + i sin t) = R (cos kt + i sin kt) R (cos t + i sin t)



Rewrite the above as follows [ R (cos t + i sin t) ]



k+1

=R

k+1

[ (cos kt cos t - sin kt sin t) + i (sin kt cos t + cos kt sin t) ]

Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t - sin kt sin t) and (sin kt cos t + cos kt sin t) as follows (cos kt cos t - sin kt sin t) = cos(kt + t) = cos(k + 1)t (sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t



Substitute the above into the last equation to obtain [ R (cos t + i sin t) ]



k+1

=R

k+1

[ cos (k + 1)t + sin(k + 1)t ]

It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1. Geometric Sequences Problems with Solutions

Problems and exercises involving geometric sequences, along with detailed solutions and answers, are presented.

REVIEW OF GEOMETRIC SEQUENCES The sequence shown below 2 , 8 , 32 , 128 , ...

has been obtained starting from 2 and multiplying each term by 4. 2 is the first term of the sequence and 4 is the common ratio. 8 = 2 * 4 or 8 / 2 = 4 32 = 8 * 4 or 32 / 8 = 4 128 = 32 * 4 or 128 / 32 = 4 and so on. The in the sequence may also be written as follows a1 = 2 a2 = a1 * 4 = 2 * 4 a3 = a2 * 4 = 2 * 4

2

a4 = a3 * 4 = 2 * 4

3

The n th term may now be written as n-1

an = a1 * r

where a1 is the first term of the sequence and r is the common ratio which is equal to 4 in the above example.

The sum of the first n of a geometric sequence is given by n

sn = a1 + a2 + a3 + ... an = a1 (1 - r ) / (1 - r) The sum S of an infinite (n approaches infinity) geometric sequence and when |r| <1 is given by S = a1 / (1 - r)

Problem 1: Find the a2, a3, a4 and a5 of a geometric sequence if a1 = 10 and the common ratio r = - 1. Solution to Problem 1: 

Use the definition of a geometric sequence a2 = a1 * r = 10 * (-1) = - 10 a3 = a2 * r = - 10 * (-1) = 10 a4 = a3 * r = 10 * (-1) = - 10 a5 = a4 * r = - 10 * (-1) = 10

Problem 2: Find the 10 th term of a geometric sequence if a1 = 45 and the common ration r = 0.2. Solution to Problem 2: 

Use the formula an = a1 * r a10 = a1 * r

n-1

that gives the n th term to find a10 as follows

n-1

9

= 45 * 0.2 = 2.304 * 10

-5

Problem 3: Find a20 of a geometric sequence if the first few of the sequence are given by -1/2 , 1/4 , -1/8 , 1 / 16 , ... Solution to Problem 3: 

We first use the first few to find the common ratio r = a2 / a1 = (1/4) / (-1/2) = -1/2 r = a3 / a2 = (-1/8) / (1/4) = -1/2 r = a4 / a3 = (1/16) / (-1/8) = -1/2



The common ration r = -1/2. We now use the formula an = a1 * r as follows. a20 = a1 * r

n-1

for the n th term to find a20

20-1

= (-1/2) * (-1/2)

20-1

20

= 1 / (20 )

Problem 4: Given the a10 = 3 / 512 and a15 = 3 / 16384 of a geometric sequence, find the exact value of the term a30 of the sequence. Solution to Problem 4: 



We first use the formula for the n th term to write a10 and a15 as follows a10 = a1 * r

10-1

= 3 / 512

a15 = a1 * r

15-1

= 3 / 16384

We now divide the a10 and a15 to write a15 / a10 = (a1 * r



14

9

/ a1 * r ) = (3 / 16384) / (3 / 512)

Solve for r to obtain. 5

r = 1 / 32 which gives r = 1/2 

We now use a10 to find a1 as follows. a10 = 3 / 512 = a1(1/2)



9

Solve for a1 to obtain. a1 = 3



We now use the formula for the n th term to find a30 as follows. a30 = 3(1/2)

29

= 3 / 536870912

Problem 5: Find the sum 6 k-1 S=∑3 k=1 Solution to Problem 5: 

We first rewrite the sum S as follows S = 1 + 3 + 9 + 27 + 81 + 243 = 364



Another method is to first note that the making the sum are those of an arithmetic n sequence with a1 = 1 and r = 3 using the formula s n = a1 (1 - r ) / (1 - r) with n = 6. 6

s6 = 1 (1 - 3 ) / (1 - 3) = 364

Problem 6: Find the sum 10 i-1 S = ∑ 8*(1/4) i=1 Solution to Problem 6: 

An examination of the included in the sum are 1

2

8 , 8*((1/4) , 8*((1/4) , ... , 8*((1/4) 

9

These are the of a geometric sequence with a1 = 8 and r = 1/4 and therefore we can use the formula for the sum of the of a geometric sequence n

s10 = a1 (1 - r ) / (1 - r) 10

= 8 * (1 - (1/4) ) / (1 - 1/4) = 10.67 (rounded to 2 decimal places)

Problem 7: Write the rational number 5.31313131... as the ratio of two integers. Solution to Problem 7: 

We first write the given rational number as an infinite sum as follows 5.313131... = 5 + 0.31 + 0.0031 + 0.000031 + ....



The making 0.31 + 0.0031 + 0.000031 ... are those of a geometric sequence with a 1 = 0.31 and r = 0.01. Hence the use of the formula for an infinite sum of a geometric sequence S = a1 / (1 - r) = 0.31 / (1 - 0.01) = 0.31 / 0.99 = 31 / 99



We now write 5.313131... as follows 5.313131... = 5 + 31/99 = 526 / 99

Exercises: Answer the following questions related to geometric sequences: a) Find a20 given that a3 = 1/2 and a5 = 8 b) Find a30 given that the first few of a geometric sequence are given by -2 , 1 , -1/2 , 1/4 ... c) Find r given that a1 = 10 and a20 = 10

-18

d) write the rational number 0.9717171... as a ratio of two positive integers. Solutions to Above Exercises: 18

a) a20 = 2

28

b) a30 = 1 / 2 c) r = 0.1

d) 0.9717171... = 481/495 Arithmetic Sequences Problems with Solutions A set of problems and exercises involving arithmetic sequences, along with detailed solutions and answers, are presented. REVIEW OF ARITHMETIC SEQUENCES The formula for the n th term an of an arithmetic sequence with a common difference d and a first term a1 is given by an = a1 + (n - 1 )d The sum sn of the first n of an arithmetic sequence is defined by sn = a1 + a2 + a3 + ... + a n and is a1 is given by sn = n (a1 + an) / 2 Problem 1: The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. Find a formula for the n th term and the value of the 50 th term Solution to Problem 1: 

Use the value of the common difference d = 3 and the first term a1 = 6 in the formula for the n th term given above an = a1 + (n - 1 )d = 6 + 3 (n - 1) =3n+3



The 50 th term is found by setting n = 50 in the above formula. a50 = 3 (50) + 3 = 153

Problem 2: The first term of an arithmetic sequence is equal to 200 and the common difference is equal to -10. Find the value of the 20 th term Solution to Problem 2: 

Use the value of the common difference d = -10 and the first term a1 = 200 in the formula for the n th term given above and then apply it to the 20 th term a20 = 200 + (-10) (20 - 1 ) = 10

Problem 3: An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term. Solution to Problem 3: 

We use the n th term formula for the 6 th term, which is known, to write a6 = 52 = a1 + 10 (6 - 1 )



The above equation allows us to calculate a1. a1 = 2



Now that we know the first term and the common difference, we use the n th term formula to find the 15 th term as follows. a15 = 2 + 10 (15 - 1) = 142

Problem 4: An arithmetic sequence has a its 5 th term equal to 22 and its 15 th term equal to 62. Find its 100 th term. Solution to Problem 4: 

We use the n th term formula for the 5 th and 15 th to write a5 = a1 + (5 - 1 ) d = 22 a15 = a1 + (15 - 1 ) d = 62



We obtain a system of 2 linear equations where the unknown are a1 and d. Subtract the right and left term of the two equations to obtain 62 - 22 = 14 d - 4 d



Solve for d. d=4



Now use the value of d in one of the equations to find a1. a1 + (5 - 1 ) 4 = 22



Solve for a1 to obtain. a1 = 6



Now that we have calculated a1 and d we use them in the n th term formula to find the 100 th formula. a100 = 6 + 4 (100 - 1 )= 402

Problem 5: Find the sum of all the integers from 1 to 1000. Solution to Problem 5: 

The sequence of integers starting from 1 to 1000 is given by 1 , 2 , 3 , 4 , ... , 1000



The above sequence has 1000 . The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n of an arithmetic sequence knowing the first and last term of the sequence and the number of (see formula above). s1000 = 1000 (1 + 1000) / 2 = 500500

Problem 6: Find the sum of the first 50 even positive integers. Solution to Problem 6: 

The sequence of the first 50 even positive integers is given by 2 , 4 , 6 , ...



The above sequence has a first term equal to 2 and a common difference d = 2. We use the n th term formula to find the 50 th term a50 = 2 + 2 (50 - 1) = 100



We now the first term and last term and the number of in the sequence, we now find the sum of the first 50 s50 = 50 (2 + 100) / 2 = 2550

Problem 7: Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5. Solution to Problem 7:



The first few of a sequence of positive integers divisible by 5 is given by 5 , 10 , 15 , ...



The above sequence has a first term equal to 5 and a common difference d = 5. We need to know the rank of the term 1555. We use the formula for the n th term as follows 1555 = a1 + (n - 1 )d



Substitute a1 and d by their values 1555 = 5 + 5(n - 1 )



Solve for n to obtain n = 311



We now know that 1555 is the 311 th term, we can use the formula for the sum as follows s311 = 311 (5 + 1555) / 2 = 242580

Problem 8: Find the sum S defined by 10 S = ∑ (2n + 1 / 2) n=1 Solution to Problem 8: 

Let us first decompose this sum as follows 10 S = ∑ (2n + 1 / 2) n=1 10 10 = 2 ∑ n + ∑ (1 / 2) i=1 n=1

 

The term ∑ n is the sum of the first 10 positive integers. The 10 first positive integers make an arirhmetic sequence with first term equal to 1, it has n = 10 and its 10 th term is equal to 10. This sum is obtained using the formula sn = n (a1 + an) / 2 as follows 10(1+10)/2 = 55



The term ∑ (1 / 2) is the addition of a constant term 10 times and is given by 10(1/2) = 5



The sum S is given by



S = 2(55) + 5 = 115 Exercises: Answer the following questions related to arithmetic sequences: a) Find a20 given that a3 = 9 and a8 = 24 b) Find a30 given that the first few of an arithmetic sequence are given by 6,12,18,... c) Find d given that a1 = 10 and a20 = 466 d) Find s30 given that a10 = 28 and a20 = 58 e) Find the sum S defined by

20 S = ∑ (3n - 1 / 2) n=1  f) Find the sum S defined by 20 40 S = ∑ 0.2 n + ∑ 0.4 j n=1 j=21  Solutions to Above Exercises: a) a20 = 60 b) a30 = 180 c) d = 24 d) s30 = 1335 e) 1380 f) 286 Find Values Of Trigonometric Functions - Questions Questions with answers on how to find exact values of trigonometric functions. You might need the table of values of the trigonometric functions for special angles. In all questions below, you are required to find exact values, if possible, of given trigonometric functions. Question 1: sin ( -π / 4) Question 2: tan ( - 3 π / 4)

Question 3: sec ( 5 π / 3) Question 4: cot ( - 5 π / 3) Question 5: cos ( 25 π / 3) Question 6: csc ( 15 π / 6) Question 7: sin( 17 π / 4 ) Question 8: sec ( π / 2 ) Question 9: tan ( 5 π / 6) Question 10: csc ( 15 π / 2) Answers to Above Questions: Note: sqrt means square root. question 1: - sqrt ( 2 ) / 2 question 2: 1 question 3: 2 question 4: sqrt( 3 ) / 3 question 5: 1 / 2 question 6: 1 question 7: sqrt( 2 ) / 2 question 8: undefined question 9: - sqrt( 3 ) / 3 question 10: -1