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SIMPLE INTEREST

SI = Simple Interest

Basics —

• P R T

SI100

• A=P+SI

• A 100

P100 R T

Where

P = Principle or Sum

R = Rate

And T = Time

A = Amount

Time saving tricks

• If a sum makes ‘n’ times of itself in ‘t’ years then

Rate = (n 1) 100

t

• If a sum make ‘n’ times of itself at rate of ‘r’ then

Time = (n 1) 100

R

• If a sum become ‘m’ times in ‘t’ years then it become ‘n’ times in

Time = (n 1) t

(m 1)

• If a sum ‘P’ become ‘A’ after t years ,then if rate increases and decreases by ‘R%’ then ‘P’ becomes

A’ = P R T

A100

Say, if 500 becomes 550 in one year at a certain rate, then if rate if increased by 5% , then after one year new amount will be

A’ = (500 5 1)

550 575100

• If a sum becomes P in ‘a’ years and become Q in ‘b’ years at simple interest then

Rate = (Q–P) 100

diff. of P b and Q a

and

sum= diff. of P b and Q a

diff. of b and a

• If sum is laid on r1% for t1 years, r2% for t2 years and so on then

SI = 1 1 2 2 3 3P (r t r t r t ....)

100

• If sum is made ‘X’ times, rate is made ‘Y’ times and time is made ‘Z’ times then percentage increase or decrease in SI would be

% inc. / dec. in SI = (X × Y × Z – 1) × 100

Say, if a principle is tripled and rate is halved and time is made 4 times than what is the % increase in SI?

% increase in SI = 13 4–1 100

2

= 500%

• If SI is a/b times of Sum and numerical value

of Rate and Time is same then

Time/Rate = a

100b

COMPOUND INTEREST

BASICS

T

T

RA P 1

100

RCI P 1 1

100

• Equivalent Rate

When T = 1

Year When T = 2

Years When T=3

Years

RATE RATE RATE

2% 4.04% 6.1208%

3% 6.09% 9.2727%

4% 8016% 12.4864%

5% 10.25% 15.7625%

6% 12.36% 19.1016%

7% 14.49% 22.5043%

8% 16.64% 25.9712%

9% 18.81% 29.5029%

10% 21% 33.1%

• If time is given in fraction then, say time

= Yx

Z

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Then x

R R YA P 1 1

100 100 Z

• To find out rate in different cases

➢ When A, P, T is given

1/t

AR –1 100

P

➢ When sum become A1 in t1 time and A1 in t2 time

2 1

1

t –t2

1

AR –1 100%

A

➢ When a sum become ‘x’ times of itself in ‘t’

years

1

tR x –1 100%

➢ When a sum becomes “n1’ in ‘t1’ time and ‘n2’ in ‘t2’ time

2 1

1

t –t2

1

nR –1 100%

n

• If a sum become

‘𝑚𝑛 ’ in ‘t’ years then it becomes ‘𝑚𝑝’ in

Time = pt

n

• When difference (D) between CI and SI is

given

For two years D = 2

RP

100

➢ For three years

2

R 300 RD P

100 100

• When CI and SI of two years is given

Rate = CI–SI100%

SI of one year

• When CI and SI of second year is given

Rate = CI SI

100%SI

Sum =

2(SI)

CI SI

• Installments Amount of every installments

= 1 2

Total amount

100 100.....n times

100 R 100 R

where n is no. of installments.

PROFIT AND LOSS

Profit (P) = SP –CP

Loss (L) = CP-SP

P or L% = P or L

100CP

CP = 100SP

100 PorL%

SP = 100 PorL%CP

100

TIME SAVING METHODS

• If an item is get at x%’ of P/L at SP1 and ‘y%’

at SP2 then

1

2

SP 100 x%

SP 100 y%

(use + sign for profit and – sign for loss)

• If an item is get ‘x%’ of P/L at Selling ‘A’ items at Rs ‘R1’ and ‘y%’ P/L at selling ‘B’ items at Rs ‘R2’ then

Then,

1

2

R

A 100 x

R 100 y

B

(use + sign for profit and – sign for loss)

• If cost price of ‘X’ items of is equal to selling

price of ‘Y’ items, then

P/L = x y100%

Y

• If ‘X’ items are purchased in ‘Y’ Rs and ‘Y’ items are sold in ‘X’ Rs, then

➢ Profit% = 2 2

2

X Y100%

Y

➢ Loss% = 2 2

2

Y X100%

Y

• If ‘x’ items are purchased in ‘a’ Rs and ‘y’ items are sold in ‘b’ Rs

➢ Profit = bx – ay 100%ay

➢ Loss = ay –bx 100%ay

• In case of dishonest shopkeeper

P or L % =

Error made by shopkeeper

Original Value Error made by shopkeeper

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If two items are of same CP, one is sold at a

profit of ‘a%’ and other is sold at a loss of ‘b%’ then

Overall P or L% = (a b)%

2

• NOTE: If two items are sold at same SP, one

at ‘x%’ profit and other at ‘x%’ loss then always Loss will occur

Overall L% = 2x

%100

And if two items are having same CP and at ‘x%’ profit and other at ‘x%’ loss then

Overall P or L% =0 %

But: If two items are sold at same SP, one at

‘x%’ profit and other at ‘y%’ loss then

remember net Profit or Loss % is

xyx y

100

In this case, net Profit or Loss %

= 100(x – y) 2xy

200 x y

• If two items are have total CP = A and one

item is sold at ‘P%’ profit and other is sold at ‘L%’ loss then

➢ CP of item sold at profit = LA

P L

➢ CP of item sold at loss = PA

P L

• If two items have a total CP = A and have same SP and one is sold at ‘x%’ profit and other is sold at ‘y%’ loss then,

➢ CP of item sold on profit

= 100 yA

200 x y

➢ CP of item sold on loss = 100 xA

200 x y

• If on selling an item at Rs X and profit percentage is numerically equal to CP then

CP = 10 x 25–50

• If on selling an item at Rs ‘X’ and profit is equal to loss occurs on selling it at Rs ‘Y’ , then

CP = x y

2

PERCENTAGE

Fraction values

Percentage Fraction values

Percentage

1/2 50 1/3 133

3

1/4 25 1/5 20

1/6 216

3

1/7 214

7

1/8 112

2

1/10 10

1/11 19

11

1/12 18

3

1/13 97

13

1/14 17

7

1/15 26

3

1/16 16

4

1/20 5 1/24 14

6

1/25 4 1/30 13

3

1/40 12

2

1/50 2

1/50 2 5/6 183

3

3/8 137

2

5/8 162

2

4/7 157

7

5/7 371

7

• ‘P%’ increase/decrease while making the

expenditure same

Required inc./dec. in consumption

= P

100100 P

• If value of an item is increased from ‘P’ Rs/Kg to ‘Q’ Rs/Kg, then for considering the expenditure same, the required decrease in consumption will be

= Q–P 100Q

• When we require equivalent percentage of x % and y%

Req. = xyx y

100

• If a person spends ‘A%’ on first thing, ‘B%’ of remaining on another thing and ‘C%’ of

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remaining on other thing and left with Rs ‘x’,

then his income will be

INCOME =

100 100 100

100 A 100 B 100 C

• If cost of sugar is increased by 20% and

hence a person purchase 5kg less sugar on Rs 500. Then what is the original and increased price of sugar?

Short Trick: Inc. / dec. price (P’)

= (price given) (% inc. / dec.)

100 (Quantity given)

Original Price (P) = P’ ×

100

100 % inc. / dec.

• Use + for profit and – for loss.

Increased price = 500 20

20 Rs/Kg100 5

Original price = 100

20 16.67Rs/Kg100 20

• If a candidate gets ‘P%’ of votes and get lost by ‘a’ votes then

Total no. of votes will be = a

5050 P

• If price of sugar is increased by 20% and a lady can only increase her expenditure by 10%, then by what % consumption will be

reduced so that expenditure will not exceed further,

Short Trick:

Dec in consumption :- a–b100

100 a

Dec in consumption = 20–10100

100 20

= 8.33 %

• In a mixture of salt of 8 litres, 40% is salt, if 2

litres water is vaporized then how much salt will left now?

Short Trick:-

given value %givenLeft salt

given value added or extracted value

(+ is used for added and – is used for extracted)

Left salt = 8 40

53.33%8 2

• If income of ‘A’ is x% of ‘B’ then what ‘B’s income is what % of ‘A’s income

100 100%

x

• If initial population/rent is ‘A’ and it is increased/decreased by ‘x%’ in first year and ‘y%’ in second year so on for ‘n’ years then final population/rent will be

Final value = initial value × x y1 1

100 100

Speed, Time and distance

DistanceSpeed

Time (m/sec)

• Note that in this equation if one thing is constant and other two are varying then we can easily solve questions by simple proportionate method or by applying ratio.

For Example:

If a car covers a distance of 144 km in 4 hrs then how much distance it will cover in 9 hrs.

Here speed is constant and distance and time varies So,

By simple Ratio method

1 2

1 2

2

2

D D

T T

D144

4 9

D 324km

Like this way we can solve countless examples in easy way.

• When relation between V1 (speed), V2 (speed),

D (distance) and ∆T is given

1 2

1 2

V VD T

V V

(where T is Time difference).

➢ (Note here we cannot apply simple ratio method because here all D,V, and T are variable.)

Types of question on which above formula

implies are

➢ A man goes with a speed of ‘x’ and reaches the shop ‘a’ minutes late. If he walks with a speed of ‘y’ then he reaches the shop ‘b’ minutes earlier. Find his distance to shop?

➢ A boy goes to school at a distance of ‘D’ kms with a speed of ‘x’. find the time he will save if

goes with the speed of ‘y’?(y>x)

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➢ If a boy covers a certain distance ‘D’ in ‘t1’

time. Then with what certain speed he will cover the distance in ‘t2’ time?

• Average speed

➢ When two speeds x and y are given

Vavg = 2xy

x y

➢ When three speeds x, y and z are given

Vavg = 3xyz

xy yz zx

➢ When more than three speeds are given

Vavg = Total distance

Total time

• If a person walks with a speed of th

x

y

of its

original speed and reaches the destination by

‘t’ minutes late, then

Original Speed = xt

x y

• If a speed of a bus without rest is ‘x’ km/hr. if it travels with delays/rest then its speed is ‘y’. then find out how much it delayed every hour?

delay or rest per hour

= difference60 minutes

Original speed

OR x y

60x

• If a vehicle goes for a certain distance by speed of ‘x’ and return the same distance with a speed of ‘y’ and take a total time of ‘t’ hours then

➢ Total Distance = 2xyt

x y

• If a person take ‘x’ hrs by waking to a place and return by a horse. And take ‘y’ hrs when goes and return both by horse. Then

➢ Time taken for going and returning the distance through walking = x+y hrs

➢ Time taken for going and returning the

distance through horse = x-y hrs

• If A starts from a point P with a speed of ‘x’ and B starts from a point Q with a speed of ‘y’, they meet at a point C and after meeting at C, A take t1 time to reach Q and B take t2 time to reach P. Then ratio of their speed

1

22

1

tx

y t

• First train leaves P at ‘a’ o’clock and reach Q

at ‘b’ o’clock and another train starts from Q at ‘x’ o’clock and reach P at ‘y’ o’clock. Then both trains will meet at

Time of meeting the trains = (b a)(y a)a

(b a) (y x)

• Two guns shots are fired from one place with a time interval of 16 minutes. But a man in the train which is moving opposite to bullets hears the shots with a time interval of 15 minutes. If speed of sound is 300m/sec , then what is the speed of train?

Short Trick:

Speedtrain × Timetrain = Speedsound × ∆T (where

∆T is time difference)

S × 15 = 300 × (16-15)

S = 20 m/sec OR 72 kmph

• train crosses a man (standing on platform) in 40 sec and a train of same length, coming from opposite to the direction of first train crosses the man in 60 sec. find the time in which the trains cross each other?

Short Trick:

1 2

1 2

2t tT

t t

2 40 60T 48sec

40 60

➢ Similarly, if both trains are moving in same

direction then time will be,

1 2

1 2

2t tT

t t

A train crosses two man going in same direction with speeds of 8m/sec 12m/sec in 8sec and 10sec respectively. Find speed of

train?

Short Trick:

1 1 2 2

1 2

diff. of (v t v t )Speed

t t

(where, v is speed and t is time)

12 10 8 8Speed 28 m/sec

10 8

Average and Age Relation

• Sum of first ‘n’ natural numbers = n(n 1)

2

• Average of first ‘n’ natural numbers = (n 1)

2

• Sum of square of first ‘n’ natural no

= n(n 1)(2n 1)

6

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• Average of square of first n natural no.

= (n 1)(2n 1)

6

• Sum of cubes of first n natural no.

= 2

n(n 1)

2

• Average of cubes of first n natural no.

= 2

(n 1)n

2

• Sum of first n even no. = n(n+1)

• Average of first n even no. = (n+1)

• Sum of first n odd no. = 2n

• Average of first n odd no. = n

• Sum of squares of first n odd/even no.

= N(N 1)(N 2)

6

Note here N = Last term not no. of terms.

For example:

12 + 32 + 52 = 5 6 7

6

= 35 OR 22 + 42 + 62

= 6 7 8

6

= 56

• Average of any uniform series will always be

= first term last term

2

For example:

5,10,15,20,25,30,35,40

Average of above series will be

= (5 10 15 20 25 30 35 40)

22.58

OR

Average = first term last term

2

= 5 40

2

= 22.5

• Average of first ‘n’ multiples of ‘k’ = (n 1)

k2

• Average of first ‘n’ whole numbers = (n 1)

2

Different types of questions and their speedy methods.

• If average of five consecutive even no. is 20, then find the max and min value?

Short Trick:

For ‘n’ consecutive even/odd no.

Max value = avg +(n - 1)

Min value = avg - (n-1)

Max value = 20 + (5-1) = 24

Min value = 20 – (5-1) = 16

• Average weight of 15 students is 30 kg. if teacher is also included then average wt. will increase by 2kgs.then find the weight of

teacher? Short Trick:

Teacher’s age =15 × 2 + (30+2) = 62 years

• Average of 15 students is 50 kgs. On coming of a new person the average weight decreases by 4kgs. Find the weight of new member?

Wt. of new member = 15 × 4 - (50 - 4) = 14kgs

NOTE – When In a group one person comes or

goes from group, then the age/weight of that Person will be

New person weight = original no. of member × (increase or decrease in average) new average

For “Included/added/increased” use + sign

For “excluded/goes/decreased” use - sign

• When two or more persons added or left the group then the avg. age/wt of the

added/reduced members will be

Short Trick:

avg. age/weight of the

added / reduced members

new averageoriginalno.of member 

increase or decrease in average no.of newly added /excluded members

no.of newly added /excluded members

For “Included/added/increased” use + sign

For “excluded/goes/decreased” use - sign

For Example: If avg. of 13 students of a group is 40kg. when two new members added

to the group then avg. of the group is

increased by 4kgs. Find avg. of two new members?

Avg. of new members

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= (13 4) (40 4) (2)

70kg.2

• When one person of a group is replaced by other, then wt/age of new member

Short Trick:

New person’s age = (total no. of members of group ×

increase/decrease in avg) (age of replaced person)

For “increase in average ” use + sign

For “decrease in average” use - sign

For Example: In a class of 8 members average Wt. of class in increased by 2kgs when one boy of wt. 30kg is replaced by a new boy? Find out the Wt. of new boy?

Wt. of new boy = (8 × 2) + 30 = 46kg

• When all the numbers of a data is

increased/decreased/multiplied/divided by a number and new avg. is asked then we simply increased/decreased/multiplied/divided the old Avg. with that number.

For Example: Average of 12 nos. is 20. If every number is divided by 2 then find the new avg. of 12 numbers?

New avg. of 12 numbers = 20

2=10

• When avg. of consecutive odd/even nos. is given, ie., always remember that consecutive odd/even nos. form uniform series and we

know for a uniform series the average is always the middle term, hence this concept

will be used to find out the consecutive odd nos.?

For Example: If avg. of five consecutive odd nos. is 15. Then find out the greatest number?

Say, five consecutive odd numbers = A, B, C, D, E

And we know by above concept

That the avg. of above five numbers will be C (i.e. middle term of the series)

Hence, C = 15,

So, E=15 +4 =19 (difference between two

consecutive odd nos. = 2)

• When from a series/group/class some number

is miss-spelt/read and we have to calculate the corrected average

Short Trick:

true value false valueCorr. avg. (old avg.)

Total no.

For example:

Average of 50 nos. is 32. But it was found that 89 has been read in place of 39. Then

Find out the new avg.?

89 39Corr. avg. (32) 31

50

AGE RELATIONS

• If age of a person = x years

His age after t years = x+t years

His age before t years = x - t years

• If age of a person before t1 years is x years,

and his age after t2 years will be = x + t1 + t2

• If age of a person after t1 years is x years, and his age before t2 years will be = x - t1 - t2

• If ratio of a:b = 3:5 and after 12 years ratio of their ages will be 3:4. Find present ages of

A and B?

In this type of questions best method is to equate ratios. As the increment in both ratios is 10 years hence we will first make the difference of ratios equal

Note-But originally there is a increase of 12

years, so we have to multiply +3 with 4

therefore, value of 1 ratio is 4

Hence, multiply 4 in each ratio to get original values

3 4 12 6 4 24

5 4 20 8 4 32

Original ages of Ages of A and B

A and B after 12 years

If A is as much younger than B as he is older than C then,

Short Trick:

A : B : C : = 1:2:3

Age of A = (sum of ages of B + C)

• If Bs present age is ‘x’ times of the age of A before ‘t’ years, then

Age of A tB x

x 1

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• If difference of ages of two persons A and B is

given and ratio of ages after/before ‘t’ years become a:b, then

A’s age after/before ‘t’ years

= a(diff. of ages)

diff. of a and b

B’s age after/before ‘t’ years

= b(diff. of ages)

diff. of a and b

• If sum of ages of A and B is given as ‘y’ and t years before A’s age was ‘n’ times of B’s age then

Present age of y (n 1)B

n 1

Power and Indices

• 𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛

• n m

m na a

• 𝑎−𝑎 = 1/𝑎𝑎

• √𝑎𝑛

= 𝑎1/𝑛

• 𝑎𝑚/𝑛 = ( 𝑎1/𝑛)𝑚 = (𝑎𝑚)1/𝑛

• 𝑎𝑚 / 𝑎𝑛 = 𝑎𝑚−𝑛

• 𝑎0 = 1

• √𝑎𝑚𝑛 = (𝑎𝑚/𝑛)

• ( √𝑎𝑛 )n = a

• √𝑎𝑛

∗ √𝑏𝑛

= √𝑎𝑏𝑛

• a b a b 2 ab

a ba b

• a b a b 2 ab

a ba b

• a/√𝑥 = a√𝑥/x

• a/(√𝑥 + √𝑦) = a(√𝑥 − √𝑦)x-y

• √𝑥 + √𝑦 + √𝑥 − √𝑦 = 2xy

√𝑥 − √𝑦 √𝑥 + √𝑦 x-y

• √ 𝑥 √ 𝑥 √𝑥 √𝑥 … … … … … . 𝑢𝑝𝑡𝑜 𝑛 𝑡𝑒𝑟𝑚𝑠

= 𝑥(2𝑛

− 1)/2𝑛

Note - (2𝑛 − 1)/2𝑛 is in power of x

• √ 𝑥 √ 𝑥 √𝑥 √𝑥 … … … … … . ∞ = x

• In all other forms just assume the complex term as say ‘x’ and proceed.

For example: Find the value of

√ 2 √4 √2 √4

3 … … … . . ∞ 3

Say

𝑥 =√ 2 √4 √2 √4

3 … … … . . ∞ 3

…(i)

𝑥2 = 2√4 √2 √43 … … … . . ∞

3

(𝑥2)3 = 8 ∗ 4 ∗ 𝑥 from eq. (i)

𝑥6 = 32 𝑥

So, 𝑥5 = 32

𝑥 = 2.

• in the expression

“√ 𝑥 + √ 𝑥 + √𝑥 + √𝑥 … … … … … . ∞ “ if

x=x*(x+1),

then

answer will always be = (x+1)

For example: Find value of

√ 12 + √ 12 + √12 + √12 … . . ∞

here, 12 = 3*4, answer will be 4

• Similarly, in the expression

“√ 𝑥 − √ 𝑥 − √𝑥 − √𝑥 … … … … … . ∞ “

if x=x*(x+1),

then answer will always be = (x-1)

For example: Find value of

√ 56 − √ 56 − √56 − √56 … . . ∞

here, 56=8*7, answer will be 7

• If not so the case

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For example: √ 15 + √ 15 + √15 + … . . ∞ ,

in this type of questions where no exact form is appearing then assume the identity as ‘x’ and solve by quadratic equation

Let x =√ 15 + √ 15 + √15 + … . . ∞

𝑥2 = 15 + x

𝑥2 -x- 15 = 0

Now solve the quadratic equation by below formula

2b b – 4acX

2a

• If in a group of n persons every man

handshake with other. Then what is the total no. of handshakes have made?

In this type of questions.

Total no. of handshakes = n(n-1)

• If in a party of n persons, every person gives gift to another person. Then find out the total no. of gifts are given in the party?

This type of questions is different from above one because here every person give gift to other person not to each other, hence here

Total no. of gifts = n(n-1)/2

Series

1. Arithematic Progression

It is a progression with constant inc. /dec. ie. 3,6,9,12….

➢ General representation of an AP is

a, a + d, a+2d, a+3d……………

➢ Sum of and A.P = 𝑛

2{2a+(n-1)d}

Or

= first term last term(no. of terms)

2

Nth term of A.P Tn = a +(n-1)d

➢ If a, b, c are in AP then b is the arithmetic

mean between a and c.

so, a c

b2

➢ If a, a1, a2, a3 ……..an, b are in AP, then we can say that a1, a2, a3 ……..an are n arithmetic

means between a and b.

➢ If each term of an AP is increased, decreased, multiplied and divided by a same constant

non zero no. then resulting series will also be

a AP.

➢ In an AP sum of terms equidistance from beginning and end will be constant.

➢ To solve problems related to AP, one must assume following terms for easy and speedy solving

3 terms : (a-d),(a),(a+d)

4 terms : (a-3d),(a-d),(a+d),(a+3d)

5 terms : (a-2d),(a-d),(a),(a+d),(a+2d)

Where a= first term

d= common difference

n= no. of terms

l=last term

2. Geometric Progression

It is a series of non zero nos. in which ratio of any term and its preceding term is always

constant.

➢ General representation of a GP is : a, ar, ar2, ar3 ….

Where a is first term and r is common ratio

➢ If a, b, c are in GP then, 𝑏2 = ac

➢ nth term of a GP is: Tn = arn–1

➢ sum of first n terms of a GP is

n

n

a(r –1)(if r 1)

(r –1)sum

a(1– r )(if r 1)

(1– r)

➢ Sum of an infinite GP is

sum = a

1 r

➢ if a, b, c are in GP then b is called as

Geometric Mean(GM) between a and c.

𝑏2=ac

Note if a and c are of opposite sign then their GM doesn’t exist.

➢ Taking terms in GP

3 terms: a/r,a,ar,

5 terns : a/𝑟2,a/r,a,ar,ar2

➢ If a, b ,c are in GP then a b a

b c b

➢ In a GP Product of terms equidistance from beginning and end is remain constant.

3. Harmonic Progression

a1, a2, a3…… will be in HP if

1 2 3

1 1 1, ,

a a a ……

are in AP.

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i.e. 1/2, 1/6, 1/10 ……

Hence

➢ General term of a HP =

1 1 1, , ...

a (a d) (a 2d)

➢ If a, b, c are in HP then, 2 1 1

b a c

➢ Nth term of a HP is: Tn = 1

a (n 1)d

➢ If a, b, c are in HP, then b is called as harmonic mean(HM) between a and c.

In this case, 2ac

ba c

➢ If a, a1, a2,…………..an , b are in HP, then a1,

a2,……..an are n harmonic means between a and b.

Relationship between AM, GM and HM

➢ 2(GM) AM HM

And, AM GM HM

There properties are extremely useful in getting max. and min. values.

Miscellaneous Some useful data to remember to save time

number square cube 4th power 5th power sq. root cube root

1 1 1 1 1 1 1

2 4 8 16 32 1.414 1.260

3 9 27 81 243 1.732 1.442

4 16 64 256 1024 2.000 1.587

5 25 125 625 3125 2.236 1.710

6 36 216 1296 7776 2.449 1.817

7 49 343 2401 16807 2.646 1.913

8 64 512 4096 32768 2.828 2.000

9 81 729 6561 59049 3.000 2.080

10 100 1000 10000 100000 3.162 2.154

11 121 1331 14641 3.317 2.224

12 144 1728 20736 3.464 2.289

13 169 2197 28561 3.606 2.351

14 196 2744 38416 3.742 2.410

15 225 3375 50625 3.873 2.466

16 256 4096 65536 4.000 2.520

17 289 4913 83521 4.123 2.571

18 324 5832 104976 4.243 2.621

19 361 6859 130321 4.359 2.668

20 400 8000 160000 4.472 2.714

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21 441 9261 4.583 2.759

22 484 10648 4.690 2.802

23 529 12167 4.796 2.844

24 576 13824 4.899 2.884

25 625 15625 5.000 2.924

26 676 17576

27 729 19683

28 784 21952

29 841 24389

30 900 27000

31 961 29791

32 1024 32768

33 1089 35937

34 1156 39304

35 1225 42875

36 1296 46656

37 1369 50653

38 1444 54872

39 1521 59319

40 1600 64000

41 1681 68921

42 1764 74088

43 1849 79507

44 1936 85184

45 2025 91125

46 2116 97336

47 2209 103823

48 2304 110592

49 2401 117649

50 2500 125000

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