find the sum of the two middlemost terms of the AP: -4/3, -1 , -2/3, .... , 4 1/3.

A sum of Rs 2700 is to be given in the form of 63 prizes. If the prize is of either Rs 100 or Rs 25, find the number of prizes of each type. Can anyone help me please that how to do it?

there are 20 terms in a an as .sum of the first and last term is  88.

a)if the 10 th term is 42,what is the 11 th term?

As we made triangles using dots, assume that squares are made using dots. Find the number of dots needed for each square.Write it as a sequence.

In an arithmetic sequence the sumof 1stnine terms 279and the sum of the 1st twenty terms is 1280.then (a)question. What is the fifth term of the sequence. (b)questions.what is the sixtenth term of the sequence. (c)question.write the sequence

Sum of the area of two squares is 468centimetre square if the difference of their perimeter is 24meter find the side of the square

Sum of the area of two squares is 468centimetre square if the difference of their perimeter is 24meter find the side of the square

Sum of the digits of a two-digit numberis9.When we interchange the digits,it is found that the resulting new number is greater than the original number by 27.What is the two digit number.

Sum of the digits of a two-digit numberis9.When we interchange the digits,it is found that the resulting new number is greater than the original number by 27.What is the two digit number.

SUM OF ALL ANGLES OF A TRIANGLE GIVES 180. PROVE

sum of  and

## Write the sum of and .

= 0.777...

x  = 0.777

10x  =  7.777

10 x – x  = ( 7.777.... ) – ( 0.777....)

9 x  =  7

## Give two rational numbers whose  : (1)  Difference is a rational number (2)  Sum is a rational number (3)  Product is a rational number (4)  Division is a rational  number               Justify also.

Any example and verification of example

product          =       4/5 x 9/2 = 36/10 (Rational number)

Division         =       9/2   4/5 = 45/8  ( Rational number)

## Given two rational numbers whose (1) difference is a rational number (2) sum is a rational number (3) product is a rational number (4) division is a rational number

Any example and verification of example

let m = 4/5, n= 9/2

## A student amit of class IX is unable to write in his examination, due to fracture in his arm. Akhil a student of a class VI writes for him. The sum of their ages is  25 years. (i) Write a linear equation for the above situation and represent it graphically.  (ii) Find the age of of Akhil from the graph, when age of Amit is 14 years.

Let  Age of Amit = x years

Age of Akhil =  Y years

(i) According to the question the linear equation  for above situation is

y = 25-x

 X 0 10 15 Y 25 15 10 (ii) From the graph when Amit's age = 14  years, then Akhil's age = 11 years.

## Show that  is an irrational number.

Let    be a rational number, which can be put in the form  , where   , a and b   are co-prime

= rational

But, we know that  is an irrational number.

Thus, our assumption is wrong.

Hence, 5 is an irrational number.

## Prove that is an irrational number

Let  be a rational number

Where p and q are co-prime integers and

is divisible by 2.

P is divisible by 2.

Let p = 2r for some positive integer r

is divisible by 2.

q is divisible by 2.

From (i) and (ii), p and q are divisible by 2, which contradicts the fact that p and q are co-primes.

Hence, our assumption is false.

So,  is irrational

## Prove that   is an irrational number.

Let   be a rational number

(a and b are integers and co-primes )

On squaring both the sides,

is divisible by 3

is divisible by 3

We can write  a = 3c for some integer

is divisible by 3

b is divisible by 3.

From equations (i) ad (ii),  we get 3 as a factor of "a"  and  "b"

Which is contradicting the fact that a and b are co-primes. Hence our assumption that  is a rational number, is false. So   is an irrational number.

## if p is a prime number, then prove that  is irrational.

Let p be a prime number and if possible, let  be rational.

Let    where m and n are  integers having no

common factor other than 1 and

Then,

Squaring on both sides, we get

P divides  m2 and p  divides m . [ p divides pn2 ]

[ P is prime and p divides m2 p divides m]

Let              m = pq  for some integer q

on putting  m = pq  [in eq.(i) we get ]

pn2 = p2q2

n2 = pq2

p divides n2

and p divides n.

[ p is prime and p divides n p divides n ]

Thus p is a common factor of m and n but this contraficts  the fact that m and n have no common factor other than 1.

The contradiction arises by assuming that is a rational .

Hence,  is irrational.

## Prove that    is an irrational number. Hence show that 7+2  is also an irrational number

If possible let   be a rational number.

(i)   , where a  and b are integers and co -primes

squaring both sides, we have

is divisible by 3

a is divisible by 3    .............................(1)

We can write a = 3c for some c (integer)

(3c)2 = 3b2

is divisible by  3.

is divisible by 3   ...........................(2)

From eq (i) and (ii) we have,

3 is a factor a and b which is contradicting the fact that 'a' and 'b' are co-prime.

Thus our assumption that    is rational numbers is wrong.

Hence,  is an irrational number.

(ii) Let us assume to the contrary that   is a rational number.

p- 7q and  2q both are integers hence   is a rational number.

But this contradicts the fact that is is  irrational number. Hence is is an irrational number .

## Find the quadratic polynomial whose sum and product of the zeroes are  and   respectively.

According to the question,

Sum of zeroes  =

and product of zeroes  =

= (Sum of zeroes ) + product of zeroes

## If one of the zeroes of the quadratic polynomial is  is a negative of the other, find the value of 'k'.

Given

Let one zero  be ,

The other  =

Sum of zeroes =

Sum of zeroes  =

According to the equation,

Sum of zeroes  =

## If the zeroes of the polynomial    are double in value to the zeroes of  , find the value of  p and q.

Let,

Let the zeroes of polynomial be   and  then,

Sum of zeroes

According to the question, zeroes of   are  and

Sum of zeroes  =

Product of zeroes  =

and q = -6

## If the sum and product of the zeroes of the polynomial  is equal to 10 each, find the value  of 'a' and  'c'

Given, Polynomial

Let the zeroes  of   are  and  , then according to the question

Sum of zeroes,  ()  = Product of zeroes,  = 10

Now,

and

Hence

## If     and  are zeroes of the polynomial  then form a quadratic polynomial whose zeroes are    and

Given, p (x) = 3x2 -4x-7 and  and   are its zeroes.

For the new polynomial

Required quadratic polynomial = x2 - ( Sum of zeroes ) x + Product of zeroes

## Show that    and    are the zeroes of the polynomial  and verify the  relationship between  zeroes and  coefficients of the polynomial.

=1+2-3 =0

and

= 9 - 6 -3 = 0

are zeroes of polynomial

Relation between zeroes and coeff. of polynomial is verified.

## If   and    are the zeros of the polynomial    find the polynomial whose zeroes  are      and

Since  and  are the zeroes of the cubic polynomial

then

and

But required polynomial = x2 - (Sum of the zeroes)x + product of the zeroes

## Sum of the areas of two squares is 468m2. If the difference of their perimeter is  24 m, find the sides of the two squares.

Let the side of the smaller squrare be y and the side of the longer square by x, then.

4x - 4y = 24

According to the equation

Rejecting y = -18, as side can not be negative

and x = 18

## A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digit. Find the number.

Let units digits and tens digits of the two digit number be x and y

Number  is  10 y + x

According to question,

Also,

or y=2

Rejecting y = 0 as the number can not be zero.

Required number is  24.

## The sum of the squares of two consecutive natural numbers is  421. Find the numbers.

Let the first natural number = x

Second consecutive natural  number  = x+1

According to the question,

or

x = -15 or x = 14

Rejecting negative value

First number  =14

and second cosecutive number = 15

## The denominator of a fraction is two more than its numerator. If the sum of the fraction and its reciprocal  is  , find the fraction.

Let the numerator be x.

denominator is x+2.

fraction  =

## The numerator of a fraction is  3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is . find the original fraction.

Let the fraction be

By the given condition, new fraction

=

or

or

The fraction is  .

## If the sum of n terms of an A.P is 2n2+5n, then find the 4th term.

Let the sum of n terms of A.P = Sn

Given,

Now, nth term of AP = sn - sn-1

4th  term

## If the nth term of an AP  is 7-3n, find the sum of twenty five terms.

Here  n = 25, tn = 7-3n

Taking

Given  AP is 4,1, -2,...........

Here, a = 4, d = 1-4 = -3

## If Sn denotes, the sum of the first n terms of an AP prove that S12 = 3 (S8-S4)

Let a be the first term and the common difference be d.

## If the ratio of the sum of first n terms of two AP is (7n+1) : (4n+27), find the ratio of their mth terms.

Let  a, A be the first terms and d,D be the common difference of two AP's

Then according to question

Putting

## The common difference of an  AP is -2. Find its sum, if first term and 100 and last term is  -10.

Here, a = 100, d = -2, tn= -10

Using, tn = a+ (n-1) d

Here 56th  term is  - 10

Number of terms  in AP are  56

## The sum of first term of an AP is given by  determining the AP and the 12th term.

A P is  -1,5,11..............

## The minimum age of children to be eligible to participate in a painting competition is  8 years it is observed that the age of youngest boy was 8 years and the ages of rest of participant are having a common difference of  4 months. If the sum of ages of eldest participant in the painting competition.

a = 8, d = 1/3 years, Sn = 168

or n = -63

n = 16

Age of oldest participant  = a + 15 d  = 13 years

## The digits of a positive number of three digits are in AP and their sum is  15. The number obtained by reserving the digits is  594 less than the original number find the number.

Let the three digits be  a -d, a , a+d

Sum  = a-d+a+a+d = 3a=15 given

the  three digits are  5 -d, 5,5+d.

Original number = 100 (5-d) +10 x 5 + 1(5+d)

= 555+99d

Revered number = 100 (5+d)+10 x 5+1 (5-d)

= 555+99 d

According to question ,

(555-99d) - (555+99d)  = 594

-198 d = 594

The three digits are

5-(-3),5,5+(-3)

8,5 and 2

Original number is  8 x 100 + 5x 10 +2 x 1 = 852

## Find the sum of the two digit numbers divisible by 6.

Series of two digits numbers which are divisible by 6 is :

12, 18, 24..........................., 96

Here, a = 12, d =18-12 = 6, tn=96

tn = a+(n-1) d

## To draw a pair of tangent to a circle which are inclined to each other at an angle of  300, it is required to draw tangents at endpoints of two radii of the circle, what will be the angles between them?

Angles between the radii = 1800-300 = 1500

(Since the sum of opposite angles = 1800).

## Two circular beads of different size are joined together such that the distance between their centres is 14 cm. The sum of their areas is 130 cm2. Find the radius of each bead.

Let the radii of the circles are  r1 cm and r2 cm

And, sum of their areas

From (i) and (ii)

## In fig., two circular flower beds have been shown on two sides of a square lawn ABCD  of sides 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals  of the square lawn, find the sum of the areas of the lawn and flower beds.    3 Total area  = Area of  sector OAB + Area of sector ODC +Area of    Area of

## Two different dice are tossed together. Find the probability. (i) That the number on each die is even  (ii) That the sum of  numbers appearing on the two dice  is 5.

(i)  Even numbers  occur is (2,2) (2,4),  (2,6), (4,2), (4,4) (4,6), (6,2) (6,4), (6,6)

P  ( number of ech die is even ) =

(ii)  Sum of numbers  is  5 in (1,4) (2,3) (3,2) (4,1)

## Find the mean of the following distribution : class interval            0-6     6-12     12-18     18-24     24-30 frequency                 5         4            1           6          4

xi                                 fi                                 xifi

3                                 5                                  15

9                                 4                                  36

15                                 1                                  15

21                                 6                                126

27                                 4                                108

Total

## The mean of the following frequency distribution is  25. Find the value of p. Class interval                  0-10              10-20               20-30              30-40            40-50 Frequency                        5                   6                    10                     6                  p

Class - Interval                       Midxi                    fi                    fixi

0-10                                   5                       4                    20

10-20                                15                       6                     90

20-30                                25                     10                   250

30-40                                35                       6                   210

40-50                                45                       p                   45p

26+p               570 + 45p

650 +25 p = 570 +45 p

650 -570= 45p -25p

p = 4

## In the given figure, AB is a diameter of the circle with centre O. if AC and BD are perpendicular on  a line PQ and BD meets the circle at E, then prove that AC = ED. Proof:

=         (semi-circle)

(sum of angles of a quad.)

each angle

is a rectangle     AC = ED.

## Find the mean of the following distribution : Height (In cm)              Less than 75            Less than 100            Less than 125           less than 150       Less than 175          No. of students                    5                              11                           14                             18                       21                             Height (in cm )             Less than  200         Less than 225          Less than 250             Less than 275        Less than  300  No. of students                   28                           33                             37                               45                       50

By short-cut Method  (any method )  For making correct table  A = 187.5

Mean  = A +

Mean  = 187.5 +

Alternative method  :

By short cut method  (any method)  A = 187.5

Class- interval Height (in cm )             Frequency                   xi                                fiui

50-75                                       5                         62.5                           -5                      -25

75-100                                       6                         87.5                           -4                      -24

100-125                                       3                        112.5                          -3                        -9

125-150                                       4                        137.5                          -2                        -8

150-175                                        3                        162.5                          -1                        -3

175-200                                        7                        187.5                           0                         0

200-225                                         5                        212.5                           1                         5

225-250                                        4                         237.5                           2                         8

250-275                                        8                         262.5                           3                       24

275-300                                         5                         287.5                           4                       20

Mean

Mean  =

## If the mean of the following is  14.7, find the value of p and q. Class                  0-6      6-12      12-18      18-24      24-30      30-36      36-42      Total Frequency           10         p           4             7            q            4             1          40

xi                                        fi                                               xifi

3                                        10                                              30

9                                        p                                               9p

15                                       4                                                60

21                                       7                                               147

27                                       q                                               27q

33                                        4                                              132

39                                         1                                               39

Total

26 + p +q = 40

Substracting eq (i)  and (ii),

Putting this value  q in eq (i),

P  = 14-q = 14 -3 = 11

p = 11, q = 3.

## Find the mean and mode of the followin frequency distribution : classes                  0-10           10-20          20-30           30-40           40-50           50-60           60-70 Frequency                3                8                10                15                7                 4                3

Class interval                             xi                           fi                                  xifi

0-10                                      5                            3                                  15

10-20                                   15                            8                                 120

20-30                                   25                          10                                 250

30-40                                   35                          15                                 525

40-50                                    45                            7                                 315

50-60                                    55                           4                                  220

60-70                                     65                           3                                  195

Mean  =

Modal class  = 30-40

Mode =

=33.85

## (i) Why is Axiom 5, in the list of Euclid's axioms, considered a universal truth? (Note that the question is not about the 5th postulates) (ii) How would you rewrite Euclid's fifth postulates so that it would be easier to understand?

(i) Since it is true for things in any part of the universe so this is a universal truth.

(ii)  If the sum of the co-interior angles made by a transversal intersects two straight lines at distinct points is less than 1800, then the lines cannot be parallel.

## ABC is a right-angled triangle in which  and AB = AC. Find We have,

## In the figure AD = BD. Prove that BD (Angles opposite to equal sides are equal)

(Exterior angleis equal to the sum of interior opposite angles)

In       AB >BD                          ( Side opposite to greatest angle is the longest )

Also in    AB < AC

BD < AC.

## The angles of a quadrilateral are .  Find the smallest and largest angles of the quadrilateral.

Sum of the angles of a quadrilateral is 3600

Smallest angle =

Largest angle =

## The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Let the meassure of the angles be 3x, 5x, 9x, 13x then

[Angles of sum property of quadrilaterals]

Angles are : 360, 600, 1080, 1560

## In the given figure A,B,C, and D are four points on a circle. AC and BD intersect at E such that  and  Find the In  (Exterior angle of  a  is equal to sum of two opposite angles)

(Angles in the same segment)

## In the given figure, O is the centre of the circle and chord AC and BD intersect at P such that  and  , find the value of  . In

(Exterior angle of a    is  equal to the sum of two opposite angles)

or,

[Angle in same sagment]

## Construct a triangle whose angles are in the ratio 1:3:5 and length of sides included by first and last angles is 6 cm.

Let angles x, 3x, and 5x

(angles of sum property)

First angle  =

Second angle  =  600

Third angle    = 1000

Steps of construction:

(i)  Draw a line segment BC  = 6 cm

(ii)  At point A draw   and at point B, draw

(iii) XA and YB intersect at point C.

Thus, ABC is the required triangle. ## The volumes of two spheres are in the ratio of 64:27. Find their radii if the sum of their radii is 21 cm.

Let the radii of two spheres be

## The mean of 5 observation was calculated as 145, but it was later on detected that one observation was misread as 45 in place of 25. Find the correct mean of the observation.

Total of  all observations = 145 x 5 = 725

Correct total all observations  = 725 - 45 +25

= 705

## The mean of 10 numbers is 55. If one number is included, their mean becomes 60. Find the included number.

Given, N = 10,

Sum of 10 observations  = 10 x 55  = 550  .................(i)

When N = 11,   = 60

Sum of 11 observations = 11 x 60  = 660   ..................(ii)

observations  =  Included number

= 660-550

= 110

## The average height of 30 students is 150 cm. It was later detected that one observation 165 cm was wrongly copied as 135 cm. Find the correct mean height.

Sum of observations = 150 x 30 = 4500

Correct sum of observations = 4500 - 135 +165

= 4530

## The mean of n- observations is . If constant  ''a '' is subtracted from each observation, then show that the new mean

Given, mean of n observation =

Let observations be

If each observation is reduced by a, the observation  will be

Let the new mean be k.

The new mean is

## Find the mean  of first ten prime numbers and hence show that

First ten prime numbers are   2,3,5,7,11,13,17,19,23,29.

Now,

## A and B  are the only two outcomes of an event.   probability of (A)  = 0.72,  then what will be the probability of (B) and why?

Since, P(A) +P (B) = 1

P(B) = 1-0.72

= 0.28

Because sum of probabilities is  1.

## If sum of the zeroes of the quadratic polynomial   is  3, then find the value of k.

## If -1 is a zero of the polynomial    then calculate the other zero.

Let the other zero be k, then

## Find  a quadratic polynomial, the sum  and product of whose zeroes are 6 and 9 respectively  Hence find the zeroes.

Sum of zeroes  = 6, product of zeroes = 9

Also,

Hence zeroes are 3,3

## Form a quadratic polynomial p(x) with 3 and 2/5  as sum and product of its zeroes, respectively .

According to the question,

sum of zeroes = 3

Product of zeroes =

(Sum of zeroes) +product of zeroes

## If p,q are zeroes of polynomial  find the value of

We know that,

## Find the values of a and b , if they are the zeroes of polynomial

Product of zeroes =

then

## Write the complementary angle of

Complementary angle of

(As sum of complementary angleis )

## Write the complement of

The complement of

(As sum of complementary angles is 900)

## The angles of a quadrilateral are in the ratio 2:3:6:7 the largest angle of the quadrilateral is ..................

Let the angles of the quadrilateral be

Largest angle  =

## The angles A,B,C, and D of a quadrilateral ABCD are in the ratio 2 : 4: 5 : 7. Find the measures of these angles. what type of quadrilateral is it.  GIve reasons?

Let the measures of the angles be 2x , 4x , 5x and 7x

(Angles of sum of property )

As

is a trapezium ## ABCD  is a quadrilateral in which the bisectors of A and C meet DC produced at Y and BA produced at X respectivly. prove that , In

(By angle sum property of a  ) ....... (1)

In

(By angle sum property of a ) ........(2)

ie.,

ie., ...............(3)

But ,

.............................(4)

(Angle sum property of a quadrilateral From (3) and (4)

## In the figure, if  then  In

By angle sum property

(  angles in the same segment of a circle are equal )

## In the adjoining figure if  and   find the measures of  [Angles in the same segment of a circle]

In

[Angle sum property of triangles]

## Find the value of k, for which one root of the quadratic equation  is six times the other.

Let one root  =

Other root  = 6

Sum of roots =       ................. (i)

Product of roots

Solving (i) and (ii)

3k = k2

or k = 3

k = 0 is not possible

k = 3

## From an exteranal point P, tangents PA and PB are drawn to a circle with centre O. If 0, then find

Alternative method :

angle sum property

Also,

## The diameters of two circles with cenntre A  and B are 16 cm and 30 cm respectively. If area of a circle with centre C is equal to the sum of areas of the other two circles, then find the circumference of the circle with centre C.

Area of circle = , Let the radius of circle with centre C = R

According to question,

=

Circumference of the circle

## What is the diameter of  a circle whose area is equal to the sum of the areas of two circles of radii 40  cm and 9 cm

Area of the circle = sum of areas of two circles

Diameter of required circle = 41 x 2 = 82 cm

## If circumference of a circle is 44 cm, then what will be the area  of the circle ?

Circumference of a circle   = 44 cm

Sum of two sides of a square is 22 cm.

Area of the circle  =

= 154

## From the following frequency distribution, find the median class :  Cost of living index          1400 - 1550      1550-1700      1700-1850     1850 - 2000 Number of weeks                      8                      15                     21                   8

C.I          1400 - 1550     1550-1700     1700 - 1850       1850 - 2000

f                      8                      15                   21                       8

c.f                   8                      23                   44                        52

Median class = 1700- 1850

## In the following frequency distribution, find the median class. Height (in cm)      140 - 145      145 -150      150 - 155     155 - 160         160 - 165       165 - 170 Frequency                   5                   15                  25                 30                    15                 10

Height                   Frequency            c.f

140-145                       5                   5

145-150                      15                 20

150-155                      25                 45

155- 160                     30                 75

160-165                      15                 90

165-170                      10                 100

=

N = 100

Hence median class is 155 - 160

## The mean and median of 100 observation are  50 and 52  respectively. The value of the largest observation is  100. It was later found that it is 110 not 100. Find the true mean and median

Mean  =

Correct  ,

Correct mean  =

= 50.1

Median  will remain same ie., median  = 52

## Find the sum of the lower limit of the median class and the upper limit of the modal class: Classes       10 -2 0        20 - 30       30 - 4 0      40-50        50- 60         60-70 Frequency      1                   3                 5               9                7                3

Classes                        10 -2 0        20 - 30       30 - 4 0      40-50        50- 60         60-70

Frequency                        1                   3                 5               9                7                3

Cumlative Frequency       1                   4                 9                18             25             28

Median Class : 40- 50    Lower limit  = 40

Modal Class :  40-50       Upper limit  = 50

Their limit   = 40 + 50 = 90

## (i) Do Euclid fifth postulates imply the existence of parallel lines? Explain  (ii) Which mathematical concept is used in this problem  (iii) What is its value

(i) If a  staright line l falls on two straight lines m and n such that the sum of the interior angles on one side of l is equals to two right angles then by Euclid's fifth postulates the lines will not meet on this side of l. Next we know that the sum of the interior angles on the other side of line l will also be two right angles. Therefore, they will not meet on the other side also. so the lines m and n never meet  and are therefore, parallel.

(ii) Introduction to euclids Geometry

(iii) Universal truth.

## Prove that bisectors of pair of vertically opposite angles are in the same straight line.

Given: two lines AB and CD intersect at point O.

Also, OM and ON are the bisectors of  and  respectively

To prove :  MON is a straight line

Prove :  Since the sum of all the angles around a point O is 3600 we have

(ver :opp. is bisector of  ]

is a straight angle

## In figure m  and n are two plane mirrors perpendicular to each other. SHow that incident ray CA is parallel to reflected ray BD>

Let normals at A and B meet at P.

As mirrors are perpendicular to each other therefore    and .

So.

i.e.,

Therefore

[Angle sum property]

Also

Angle of incidence = angle of reflection

Therefore  = 900

Adding (1) and (2) we have

i.e.,

Hence

## Find the supplement of 4/3 of a right angle.

4/3 of a right angle =

(Sum of supplementary angles is1800)

Supplement of 1200= 1800  - 1200 = 600

## In  a parallelogram PQRS of the given figure the bisectors of

Given : A parallelogram PQRS in which the bisectors of LP and LQ meet SR at O.

To prove :

Now, since PQRS is a parallelogram. Therefore PS II QR

Now, PS II QR and transversal PQ intersects them.

( sum of consecutive interior angles is 1800)

(OP is  bisector of

and

Now, in

## Find the angles ABC, ADE,BCD in the adjacent figure, where 'O' is the centre of the circle.

(angle in a semi circle)

= 300

Angle sum property

Therefore

## The sum of the opposite angles of a cyclic quadrilaterals is.............

Sum of opposite angles of cyclic quadrilateral is  1800

## ABCD  is a cyclic quadrilateral in which AC and BD are its diagonals. If  and

(angles in the same segment )

In

Angle sum proerty

## ABCD is a cyclic quadrilateral in which AB II CD  find all the  remaining angles .

Since sum of the opposite pairs of angles in a  cylic quadrilateral is 1800

Hence

Again AB II CD and AD is its transversal, so

(Consecutive interior angles)

and

## Construct a right triangle with base  4.5 cm and the perimeter as 11.7 cm

Given: In right  , base BC = 4.5 cm and perimeter  = 11.7 cm

i.e.,  AB + BC = 11.7 cm or AB  + AC = 7.2 cm

Required : TO construct the  with   base BC = 4.5 cm

and sum of other two sides as  7.2 cm

Steps of construction:

(i)  Draw  BC  = 4.5 cm

(ii) Draw BY such that  < CBY  = 900

(iii)  From BY cut  off BD = 7.2 cm

(iv) Join DC.

(v) Draw the perpendicular bisector of DC intersecting BD at A.

(vi) Join AC then,  is the required triangle. ## There are 100 students in a class 40 of them are girls. The average marks of the boys in science is 75% and that of the girls is 65% . Find  the average marks of the class in science.

Total marks of boys = 60 x 75 = 4500

Total marks of girls = 40 x 65= 2600

Sum for class = 4500 + 2600 = 7100

Mean marks of the class  =

## The mean of 100 observations is 60. if one observation of 50 is replaced by 110, then what will be the new mean?

Sum of 100 observations = 60 x 100 = 6000

After replacement, sum of new obs.

= 6000 - 50 + 110 = 6060

New mean  =

## The mean of 40 observations was 160. it was detected on rechecking that the value of 165 was wrongly copied as 125. Find the correct mean.

Mean of 40 observations  = 160,

Sum of 40 observation  = 160 x 40 = 6400

New sum  = 6400 +165 -125

= 6400+40 = 6440

New mean  =

## A class consists of 50 students out of which 30 are girls. The mean of marks scored by girls in a test is  73  and that of the boy is 71.  find the mean score of the whole class.

Sum of marks obtained by 30 girls = 30 x 73 = 2190

Sum of marks obtained by 20 boys =  20 x 71 = 1420

Sum of marks of 50 students  of class = 2190 + 1420 = 3610

Mean for class  =  marks

## Find the mean of the following distribution:. Variables (x)       5     15      25      32     45 Frequency(f)       6      4       9        6        5

 x f fx 5 6 30 15 4 60 25 9 225 35 6 210 45 5 225 Total

Mean

## The mean of the monthly salary of 12 employees of a  firm is Rs. 14,500. If one more person joins the firm who gets RS. 18,400 per month, then what will the mean monthly salary now?

Mean monthly salary of 12 employees  = 14,500

Sum of monthly salary of 12 employees

= 14500 x 12

= 1,74,00

Sum of the monthly salary  of 13  employees

= 1,74,000+18,400

=1,92,400

Mean of monthly salary of 13 employees

=

## Find the mean of the following distribution : Variable (x)    5    15     25    35    45 Frequency(f)   6     4       9      6      5

 x f fx 5 6 30 15 4 60 25 9 225 35 6 210 45 5 225 Total

Mean

## Find the mean for the following data: x f fx 4 4 16 6 8 48 8 14 112 10 11 110 12 3 36 Total

Mean  =

## Find the mean for the weekly pocket money (in Rs.) using the following data x f fx 55 8 440 50 3 150 49 10 490 81 7 567 48 3 144 57 7 399 65 2 130 Total

Mean

Mean pocket money per week  = Rs. 58

## Find the mean and mode of the following data: 15,17,16,14,17,16,11,15,17,14

Arranging data in ascending order:

11,14,14,15,15, 16,16,17,17,17

Mean  =

Here n = 10  (even), median

=

Mode  = 17.

## The scores of 15 students in an examination out of 10 marks is as below : 3,9,7,5,6,3,7,6,7,4,7,7,4,8,2  Find the mean, mode and median

Writting the given data in ascending order:

2,3,3,4,4,5,6,6,7,7,7,7,7,8,9

Here, n = 15, Mean  =

Mode  = 7

Median = 8th term = 6

## Find the mean, median, mode of the following data: 41,39,48,52,41,48,36,41,37,35

Mean =

Arranging the data in increasing order, we get 35,36,37,39,41,41,41,48,48,52

Thereare 10 observations

Median= Mean of 5th and 6th observation

Mode  = 41 (Maximum freq.)

## Two dice are thrown 400 times Each time sum of two numbers appearing on the tops is noted as given in the following table: What is the probability of getting a sum  (i) 5 (ii) More than  10 (iii) Between 5 and 10

(i)

(ii) P (More than 10)         =

(iii) P ( between 5 and 10)

## The are of a trapezium is 34 cm2 and the length of one of the parallel side is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Area of the trapezium  = 1/2 x sum of parallel sides X height

34 = 1/2 (10 + x) X4

10 + x = 17

X = 7

Hence, the other side  = 7 cm.

## Find CI paid when a sum of Rs.10,000 is invested for 1 year and 3 months at 8 1/2% per annum compounded annually.

We have  R = 17/2 %

T = 1 1/4 yrs

Amount for 1 year  = 10000 (1+17/200)

=Rs. 10,850

Interset for 1st year = 10,850 - 10.000 = Rs.850

S.I for the nest 1/4 years on Rs. 10850/. =

=Rs. 230.56

Total interest = 850+230.56

=Rs.1080.56

180 degrees

## Without adding find the sum of the following 1+3+5+7

Sum of first n odd natural numbers is n2 = 42 = 16

## Draw the convex polygon having 4 and 5 sides (use angle sum of polygon with n  sided concept ) and explain?

If no other angles of a polygon area reflex angle then the polygon is called convex polygon. a. 112

## The adjacent angles of a parallelogram are (2x-4)0 and (3x-1)0. Find the measures of all angles of the parallelogram.

Sum of the adjacent angle of a parallelogram = 1800

So angles are,

## The marked price of an article is Rs. 500/- A shopkeeper gives a discount of 5%  and still makes a profit of 25%. Find the cost of that article?

First calculate the selling price after

Let us assume the cost of the article is x. Hence as per the question

## Vikas has to cut out of a diameter 11/4 cm from a rectangular aluminum strip of imenstion 8 3/4  and 1 1/4. How many full circles cikas cut? Also find out the wastage of aluminium strip?                                                        OR Deepak, Rahul and Sonu received a total of Rs. 2,016/-  as a monthly allowance from their mother such that Sonu gets 1/2  of what Deepak gets and Rahul gets  1 2/3 times Sonu's share. How much money do the three brothers get individually?

Area of rectangular strip =

Area of the circle cut  =

No. of circle cuts  =

Hence total no: of ...... circle is 35

To strip & crap =

= 1.05 cm (approx)

(b) Let as assume deepaks share is x Rs.

Sonu's share  =

Rahul's  =

Therefore

Deepak share

Sanus share

Rahul's share

## Three numbers are in the ratio of 1:2:3 and the sum of their cubes are 4500. find the numbers.

Let the nos. are x, 2x,3x

So, nos. are 5, 10 & 15

## The ratio between exterior angle and the interior angle of a regular polygon is 1:5. FInd the number of sides of the polygon.

For a regular polygon

The sum of all exterior angles of the regular polygon =

Hence

## If 31z5 is a multiple of 9, where 'z' is a digit, what is the value of z ?

Sum of the digits = 3+1+z+5 = 9+z

9+z is one of the multiples of 9 i.e., is 9+z is one of the nos.of  0,9,18,27, so on.

Since z is a digit, therefore  9+z = 9  or 18

z = 0  or  9

## Find the values of the letters A,B and C and give reasons for the steps involved Sum in ones column = A + 8 = 3

A has to be 5

(A+8 = 5+8 = 13)

Sum in ten's column= 1+4+9=14

B has to be 4

C has to be 1

## Sum of the two numbers is 95. if one exceeds the other by 15, find the number

Two numbers are   x ,x+15

## The sum of three consecutive multiples of 8 is 888. Find the multiples.

If x  is a multiple of 8, the multiples are x+8 and x+16

X + (x+8) + (x+16) = 888

x+x+8+x+16 = 888

3x+24 = 888

3x = 888-24

x = 864/3

X = 288

x+8 = 296

x +16 = 304

22

23

25

31

35

## The algebraic form for the sum of first n terms of an arithmetic sequence is 2n2 + 8n. How many consecutive terms of the sequence, starting from the first, are to be added to get  330?

Given, the sum of the first n terms of an arithmetic sequence = 2n2 + 8n

ie., 2n2 + 8n = 330    dividing by 2

ie., n + 15 = 0 or n - 11 = 0

n = -15  or n = 11 ; -15 rejected

n = 11

ie., The sum of the first n number be 11.

## There are 20 terms in an arithmetic sequence. Sum of the first and last terms is 88. a. What is the sum of the 2nd and 19th terms?  b. If the 10th term is 42, What is the 11th term?  c. What is the common difference of the sequence? d. What is the first term?

Given n =20

Sum of the first and last term  = 88

a. The sum of the 2nd and 19th terms.

ie., x2  + x19 = x1 + d + x20 - d

= x1 + x20

= 88.

b. Given 10th term 42.

x10 + x11 = (x1 + 9d) + (x20-9d)

ie., 42 + x11 = x1 + x20

x11 = x1 + x20 - 42

= 88 - 42 = 46.

Hence 11th term = 46

c. Commom difference (d) = x11 - x10 = 46 - 42 = 4

d. The first term

x1 = x1 +9d = 42 (given 10th term 42)

= 42 - 9d

= 12 - 9   4 = 6

Hence the first term  = 6

## The table below shows the members in '' Stree-sakthi Kudambasree'' sorted according to their ages. Age group                  Number of members 20-30                                      4 30-40                                      8 40-50                                      10 50-60                                       7 60-70                                      4 70-80                                      2 Total                                       35        a. if the members are arranged in increasing order of ages, the age of the member at what position is taken as the median? b. What is the assumed to be an age of the member at the 13th position? c. Find the median of the ages.

 Class Frequency (f) Cumulative frequency (cf) 20-30 4 4 30-40 8 12 40-50 10 22(N/2) 50-60 7 29 60-70 4 33 70-80 2 35 TOtal 35

a.

b. 13th position = 40 + 1/2 = 40.5

c. Median

= Lower limite of the median class,

N= Total number of frequencies,    C = Above the cf of the median class,

f = Frequency of the median class, h = class interval

N/2 = 35/2 = 17.5. Median class = 40 - 50    (N/2 = 17.5 included class)

Hence, median =

= 40 + 5.5 = 45.5.

## Consider the numbers between 100 and 300 which leave remainder 2 on division by 3. a. Which is the first number in this sequence? b. Which is the last numbers in this sequence? c. How many such numbers are there in this sequence? d. Find the sum of all numbers in the sequence?

Numbers b/w 100 and 300 which leave remainder 2 on division by 3

First term = 99 + 2 = 101.    d= 3

So, the sequence be 101, 104, 107, 110..........................299.

Last term = 300 - 1 = 299

a. The first term = 101

b. Last term  = 299.

c.

## A rope of length 40 meters is cut into two pieces and two squares are made on the floor with them. The sum of the areas enclosed is 58 square meter. a. If the length of one piece is taken as x, what is the length of the other piece? b. What are the lengths of the sides of the squares? c. Write the given fact about the area as an algebraic equation. d. What is the length of each piece?

Length of the rope = 40 cm

Let the length of the one piece = x

a. Length of the other piece = 40 - x

b. Length of the side of the square  =   and

c. The algebraic form, Given, sum of the area = 58 cm2

ie., a2 + a2 =58 cm2.

dividing throught out by 2,

ie., x2 - 40x + 800 = 464.

d)  Length of each  piec,

x = 28  or x = 12

The length of each pieces = 28 m or 12 m

## What is the sum of the angles of a 62 side polygon

Sum of angle of 62 sided polygon  =  (n-2) 180 = (62 - 2) x 1800

= 60 x 180

= 108000

## Sum of first n terms of an arithmetic sequence is 3n2 + n. Find the first term and the common difference of this sequence.

Sum of n numbers in a sequence = pn2+9n

pn2 + 9n  = 3n2 + n

here p = 3

9 = 1

Where P is the common difference = 3

and first term = P + 9 = 3 + 1 = 4

## The terms of an arithmetic sequence with common difference 4 are natural numbers.  a. If x is a term in this sequence, what is the next term? b. If the sum of reciprocals of two consecutive terms of this sequence is 4/15, find those terms. OR a. Lengths of sides of a right-angled triangle are in arithmetic sequence with common differenced. if the length of the smallest side of the triangle is x-d write the length of its other two sides. b. Show that any right-angled triangle with sides in an arithmetic sequence is similar to the right-angled triangle with sides 3,4 and 5.

Common difference = 4

a. next term after x = x  +4

b. Let the first term be x, then next  term = x + 4

15 (2x+4) = 4 (x2 + 4x)

x = 6.28 or x = -2.78

x = 6.28

x + 4 = 6.28 + 4 = 10.28

OR

a.

Common difference = d

Length of smallest side = x-d

Then other two sides = x - d + d = x

and  x + d

b. According to pythagoras theorem

Similarity consider

PR = x + d

x = 4 d

When d = 1, ABC will be formed

Hence these two triangles are similar

## Consider the arithmetic sequence 9,15,21............ a. Write the algebraic form of this sequence. b. Find the algebraic form of this sequence. c. Find the sum of terms from twenty-fifth to fiftieth of this sequence. d. Can the sum of some terms of this sequence be 2015? why?

a. Common difference = 6

Let the first term be x  then the sequence = x, x+6, x + 12, x + 18.....

b.

25th term = f + (n-1)d

=  9 + (25-1)6

= 9 +24 x 6

= 153.

c. 50th term = f + (n-1)d

= 9 + (50-1) 6

= 9 + 49 x 6

= 303

Sum of n terms of arithmetic sequence

=

=

= 2280

## Write three numbers whose average is 10. with two of them greater than 10 and one less than 10

Average = 10

Sum of three numbers = 10 x 3 = 30

Set of numbers are  7, 13, 10

## The table on the right shows 30 workers sorted according to their daily wages. a. What do we take as the mean dailly wages of the four workers earning wages between 300 rupees and 400 rupees? b. According to this. What is the total daily wages of the workers in this class? c. Making such assumptiuons. Calculate the total wages of workers in other classes also. d. Calculate the mean daily wages for the entire group ?   Wages         Workers 300-400        4 400-500       8 500-600       10 600-700       6 700-800       2

 Wages Workers Mid-value Total wages 300 - 400 4 350 1400 400 - 500 8 450 3600 500 - 600 10 550 5500 600 - 700 6 650 3900 700 - 800 2 750 1500 Total 30 15900

a) Mean daily wages of 4 workers earning wages between 300 and 400 rupees = 350

b) Total wages of workers in this class = 1400

c) Total wages of workers in other classes

class           Total wages

400 - 500     3600

500 - 600     5500

600 - 700     3900

700 - 800     1500

d) Mean daily wages =

## a. What is the sum of the inner and outer angles at the vertex of a polygon? b. If the outer angle is 20 more than thrice the inner angle, What is the measure of the inner angle?

a. The sum of the inner and outer angle = 1800

b. Let the inner angle be x

By question, outer angle = 3x + 20

ie., x + 3x + 20 = 180

4x + 20 = 180

4x = 180 - 20 = 160

x = 120/4 = 40

Hence the inner angle be 400.

## Examine the pattern: (i) Write two fractions whose sum is 1. Check whether the sum and the product of the reciprocals of these fractions are      equal. (ii) Prove that the sum of the and product of the reciprocals of two such fractions are equal.                                                                        OR   (i) If  , then find  (ii) The sum of the square of a number and 2 divided by the difference of 2 from the square of that number gives 99/97       Find the number?

(i)

Sum of the reciprocals

Product of the reciprocals

They are equal

(ii) Let the number be  and

Sum of the reciprocals

Product of the reciprocals

They are equal

OR

(i)

x = 7k, y = 5k

(ii)

Number = 14

## The sum of the digits of a two digit number is 9. The digit in the one's place is 1 more than 3 times the digit in the tens place. Find the number?

Let the number of the tens place =  x

and the digit in the ones place = y

x + y = 9 ........(1)

3x + 1 = y

3x -y = -1 ..........(2)

(1)+ (2) , 4x = 8,

From (1,) 2 + y = 9, y = 9-2 = 7

Number =27

## A rhombus with one side 3 metres and one angle 600 is given in the picture. Find the sum of the length of its diagonals correct to centimetres.  is an equilateral triangle.

BD = 3 m

In  the angles are 300, 600 and 900

So its  sides are in the ratio

CD = 3 m

OD = 1.5 m, OC = 1.5 m

AC =

=5.196 m = 5.20 m

Sum of the lengths of the diagonals  = 3 + 5.20 = 8.20 m

## (a) What are the numbers x which satisfy the equation  ? (b) What are the numbers x satisfying the equation  (c) Are there numbers x satisfying the equation  Write the reason?

(a)

The distance between x and 2+ the distance between x and 6 should be 4.

Distance between 2 and 6

So x can be anywhere between 2 and 6 including 2 and 6.

So x can be any number 2 or greater than 2 and 6 or less than 6.

(b)

The distance between the x and 2+ the distance between x and 6 should be 5.

Distance between 2 and 6=  = 4

For this distance to be 5, x should move 1/2 unit to the left of 2 and 1/2 unit to the right of 6.

Then the distance =

When x is moved  unit to the left of 2, it is .

When x is moved    unit to the right of 6, it is   .

So values of x are  and .

(C)

The sum of the distance between x and 2 and x and 6 is equal to or greater than the distance between 2 and 6. The distance between 2 and 6 is 4. So this distance cannot be less than 4.

There are no numbers for x satisfying

## Sides of a right triangle ABC are AB = 10 cm, BC = 8 cm and AC = 6 cm. Semicircles are drawn with sides of the triangle as diameters.  i. Find the areas of semicircles. ii. Prove that sum of areas of two small semicircles is equal to the area of the largest semicircle.

i) Area of the semicircle with AC as diameter sq.cm

Area of the semicircle with BC as the diameter

Area of the semicircle with AB as the diameter

ii) Sum of the areas of the two small semicircles

Area of the large semicircle is equal to the sum of the areas of the two small semicircles.

## 45 is a term in the arithmetic sequence whose common difference is 2. Check whether the sum of any 17 terms of this sequence will be 2018? why?

Since 45 is a term of an arithmetic sequence with common difference 2, all the terms in this sequence are odd numbers. The sum of 17 terms of this sequence also will be an odd number. So 2018 cannot be the sum of 17 terms of this sequence.

## find n.

(Sum of the first n odd numbers = n2)

## a) What is the sum of the first 20 natural numbers.  b) Find the sum of the first 20 terms of 4, 8, 12.... c) If 3 is added to each of term in the above sequence write down the algebraic expression of the new sequence. c)  Find the sum of the first 20 terms of the new sequence

a) Sum of the first n natural numbers =

Sum of the first 20 natural numbers  =

b) 4 + 8 + 12 + ................(20 terms)

= 4 (1 + 2 + 3 + .....+20) = 4 x 210 = 840

c) Algebraic expression of the arithmetic sequence 4, 8, 12...........= 4n

Algebraic expression of the arithmetic sequence got by adding 3 to each term of the above sequence = 4n + 3

d) 7 + 11 + 15  + ..........(20terms) + 3 x 20

= 840 + 60 = 900

## 23rd term of an arithmetic sequence is 32.35th term is 104. Then, a) What is the common difference?  b) Which is the middle term of the first 35 terms of this sequence? c) Find the sum of first 35 terms of this sequence

a) Difference between the terms = 104 - 32 = 72

Difference between the term positions = 35 - 23 = 12

72  = 12 x Common difference

Common difference = 72

b) Middle term of the first 35 terms =  term

18th term = 23rd term  - 5 common difference

= 32 - 5 x 6 = 32 - 30 = 2

c)  Sum of the first 35 terms  = 35 x middle term

= 35 x 2 = 70

## 1 2   3    4 5   6    7     8   9 a) How many numbers are there in the 30th row of this number pyramid? b) which is the last number in the 30th row? c) Which is the first number in the 30th row? d) What is the sum of all terms in the first  30 rows?

a) Sequence of the number of terms in each row = 1,3,5,7.........

Xn = 2n -1

Number of numbers in the 30th row = 2 x 30 - 1 = 60 - 1 = 59

b) Last number in the 1st row = 1

Last number in the 2nd row = 4 = 22

Last number in the 3rd row = 9 = 32

......................................................

......................................................

Last number in the 30th row  = 900 = 302

c) First number in the 30th row  = last number in the 30th row  - 58 common difference

= 900 - 58 x 1 = 842

d) Sum of all terms in the first 30 rows

= 1 + 2 + 3 + 4 +........ + 900 =

## n, 3n,5n .........  is an arithmetic sequence.  a) What is the common difference?  b) Prove that the sum of first n terms of this sequence is n3. c) Then find the sum of 15 terms of the sequence 15, 45,75.......

Arithmetic sequence = n, 3n, 5n.......

Common difference = 5n - 3n = 2n

b) Sum of the first n terms =  n + 3n + 5n + .. (n terms)

=  n (1 + 3 + 5 + .....+ 2n - 1) = n x n2 = n3

c) 15 + 45 + 75 + ...........(15 terms)

= 15 (1 + 3+5+......(15 terms)

= 15 x 152 = 3375

## Sum of the first five terms of an arithmetic sequence is 45. What is the third term? The common difference of the sequence is 4, write the first two terms. write another arithmetic sequence having the sum of the first five-term is 45.

Sum of the first terms = 45

5 x 3rd term = 45, 3rd term = 45  5 = 9

If the common difference is 4, the second term = 9 - 4 = 5

The first  term = 5 - 4 = 1, First two terms = 1, 5

If the sum of 5 terms is 45, the third term is 9. another arithmetic sequence  = 7,8,9,10,11

## a) Find the least and highest three-digit number which leave a remainder 1 on division by 9. b) How many three-digit numbers are there, which leave a remainder one on division by 9. c) Find the sum of all such numbers.

a) The least three digit number which leave a remainder 1 on division by 9 = 99 + 1 = 100

The highest three digit number like this  = 999 - 8 = 991

b) Let there  be n numbers between 100 and 991 with common difference  9

100 + (n-1) 9 = 991, 100 + 9n - 9 = 991

9n = 991 + 9 - 100 = 900, n = 900 9 = 100

c) Sum  = n/2 (first term + Last term)

= 100/2 (100+991) = 50 x 1091 = 54550

## In the circle shown, the chords AQ and BP passes through C. a. The central angle of arc AXB is  Calculate    The central angle of arc PYQ is  Find all angles of the triangle BQC. b. In the picture, prove that  is half the sum of the central angle of arc AXC and arc BYD. a)

b) Draw BC. In  = half of the central angle of arc AXC.

=  Half of the central angle of arc BYD. in a triangle the measure of an outer angle is equal to the sum of the measures of the inner angles at the other two vertices.

So,

the central angle of arc AXC +  the central angle of arc BYD

(Central angle of arc AXC + central  angle of arc BYD)

## Find the probability of getting a sum of 9, when two dice are thrown simultaneously

The two dice are thrown

Possible outcomes = 36

The sum of both faces be 9 they are,

(3,6); (6,3) (4,5) (5,4)  = 4

## Two dice are rolled simultaneously. Find the probability that the sum of numbers appearing is 10.

When two dice are thrown

Possible outcomes  = 36

If sum of both faces should be 10, they are,

## Two dice, one blue and one grey, are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is 8?

Total number of outcomes  = 36

Favourable outcomes are (2,6) (3,5) (4,4) (5,3) (6, 2) = 5

Required probability =

## Two different dice are rolled together. Find the probabity to getting: i. The sum of numbers on two dice to be 5. ii. Even numbers on both dice.

Total possible outcomes  : 36

(i) The possible outcomes are (2,3) ; (3,2) ; (1,4) ; (4,1)  = 4

Required probability P(E) =

(ii) The possible outcomes  are :

(2,2) ; (2,4) ;(2,6) ; (4,2) ; (4,4) ; (4,6) ; (6,2) ; (6,4) ; (6,6) = 9

Required probability

## Two dice are thrown at the same time. Find the probability of getting: i. Same number on both dice ii. Sum of two numbers appearing on both the dice.

(i)

(ii) P(Sum is 8) =

## In the figure, PR is the angle bisector of . Prove that . (Angle bisector)

(Vertically opposite angles

Since sum of interior angles is .

## The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of the numbers of houses proceeding the house numbered x is equal to sum of the numbers of houses following x.

The houses are numbered consecutely from 1 to 49.

1,2,3,..............x-1, x, x+1,.............49

Sum of no. of houses proceeding  x numbered houses  = Sum of the no. of folloeing x

x = 35

35, being whole no, proved that given statement is true for x = 35

## Prove that  is an irrational number. Hence show that  is alos an irrational number.

Let  be a rational number

(a, b are co-prime integers and )

Squaring,

2 divides a2

2 divides a

So we can write a = 2c for some integer c,, substitute for

This means 2 divides b2, so 2 divides b.

a and b have '2' as a common factor

But this contradicts that a, b have no commom factor other than 1.

Our assumption is wrong.

Hence,  is irrational.

Let  be rational

where a and b are integers,

is rational but  is not rational

Our assumption is wrong

is rational.

## Show that there is no positive integer n, for which  is rational.

Let us assume that there is a positive integer n for  which is rational and equal to  ,where  P and Q are positive integers ()

................(i)

................(ii)

Apply (i) + (ii) we get

.............(iii)

Apply (i) and (ii) we get

.................(iv)

From (iii) and (iv), we can say  and  both are rational because P and Q both are  rational. But it is possible only when (n+1) and (n-1)  both  are perfect squares. But they differ by 2 and two perfect square never differ by 2. So both (n+1) and (n-1) cannnot be perfect squares, hence there is no positive integer n for which  is rational.