mixture: May 2021

Monday, May 31, 2021

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3

exercise 7.2 coordinate geometry chapter 7 cbse ncert 10th mathematics

section formula, point of trisection

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3

using section formula for internal division

2 : 3 = m1 : m2

(–1, 7)=(x1,y1)

(4, –3)=(x2,y2)

required point is given

by [ { 2(4)+3(-1)}/{2+3} , {2(-3)+3(7)}/{2+3} ]

=(5/5 , 15/5) = (1,3)

2. Find the coordinates of the points of trisection of the line segment joining

(4, –1) and (-2,-3)

A=(4, –1) = (x1,y1)

B= (-2,-3)=(x2,y2)

let P ,Q be the required points [draw a rough figure.

They divide AB in the ratio 1:2, 2:1 respectively.

To find P

using section formula for internal division

1:2 = m1 : m2

P= [ {1(-2)+2(4)}/{1+2} , {1(-3)+2(-1)}/{1+2} ]

P=( 2 , (-5)/3 )

To find Q

using section formula for internal division

2:1 = m1 : m2

Q= [ {2(-2)+1(4)}/{2+1} , {2(-3)+1(-1)}/{2+1} ]

Q=[0,(-7)/3]

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ncert cbse 10th mathematics

co ordinate geometry chapter 7

exercise 7.4 optional exercise

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

solution

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

solution

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

exercise 7.3

Find the area of the triangle whose vertices are

(2, 3), (–1, 0), (2, – 4)

solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

2. In each of the following find the value of ‘k’, for which the points are collinear.

(7, –2), (5, 1), (3, k)

solution

(ii) (8, 1), (k, – 4), (2, –5)

solution

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution

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Monday, May 10, 2021

In each of the following find the value of ‘k’, for which the points are collinear. (ii) (8, 1), (k, – 4), (2, –5)

exercise 7.3 co ordinate geometry chapter 7 cbse ncert 10th mathematics

2. In each of the following find the value of ‘k’, for which the points are collinear.

(ii) (8, 1), (k, – 4), (2, –5)

collinear means that area of the triangle formed by the points is zero

(1/2) [8[-4-(-5)] +k[-5-1]+2[1-(-4)] ]=0

(1/2)[8-6k+10]=0

6k=18

k=3

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

use a rough plot of the points

let A =(– 4, – 2)

B= (– 3, – 5)

C= (3, – 2)

D= (2, 3)

draw diagonal AC and split the quadrilateral into two triangles ABC and ACD

A =(– 4, – 2)

B= (– 3, – 5)

C= (3, – 2)

area of triangle ABC =(1/2)[(-4)[-5+2]-3[-2+2]+3[-2+5]]

=(1/2)[12+0+9]=(21/2 )sq. units

A =(– 4, – 2)

C= (3, – 2)

D= (2, 3)

area of triangle ACD=(1/2)[(-4)[-2-3]+3[3+2]+2[-2+2]]

=(1/2)[20+15+0]=(35/2)sq.units.

total area of the quadrilateral =(21/2)+(35/2)=28 sq.units

=================================================

ncert cbse 10th mathematics

co ordinate geometry chapter 7

exercise 7.4 optional exercise

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

solution

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

solution

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

exercise 7.3

Find the area of the triangle whose vertices are

(2, 3), (–1, 0), (2, – 4)

solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

2. In each of the following find the value of ‘k’, for which the points are collinear.

(7, –2), (5, 1), (3, k)

solution

(ii) (8, 1), (k, – 4), (2, –5)

solution

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution

Sunday, May 9, 2021

Find the area of the triangle whose vertices are : (ii) (–5, –1), (3, –5), (5, 2)

exercise 7.3 co ordinate geometry chapter 7 cbse ncert 10th mathematics

Find the area of the triangle whose vertices are :

(ii) (–5, –1), (3, –5), (5, 2)

Area of triangle = (1/2) [ (-5)[-5-2] +3[2-(-1)]+5[-1-(-5)]

=(1/2)[35+9+20]=32 sq.units.

2. In each of the following find the value of ‘k’, for which the points are collinear.

(7, –2), (5, 1), (3, k)

collinear means that area of the triangle formed by the points is zero

(1/2) [ 7(1-k)+5[k-(-2)]+3[-2-1] ] = 0

(1/2) [7-7k+5k+10-9]=0

(1/2[8 -2k] = 0

8 -2k =0

2k=8

k=8/2

k=4

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ncert cbse 10th mathematics

co ordinate geometry chapter 7

exercise 7.4 optional exercise

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

solution

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

solution

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

exercise 7.3

Find the area of the triangle whose vertices are

(2, 3), (–1, 0), (2, – 4)

solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

2. In each of the following find the value of ‘k’, for which the points are collinear.

(7, –2), (5, 1), (3, k)

solution

Thursday, May 6, 2021

ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

exercise 7.4 optional exercise co ordinate geometry chapter 7 cbse ncert 10th mathematics

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

A(–1, –1), B(– 1, 4)

using midpoint formula

P = ( [(-1) +(-1)]/2 , [(-1)+4]/2 ) = (-1 , 3/2 )

B(– 1, 4), C(5, 4)

Q = ( [(-1)+5]/2 , [4+4]/2 ) = (2,4)

C(5, 4) D(5, – 1)

R = ( [ 5 + 5]/2 , [4+(-1)]/2 ) = (5, (3/2) )

D(5, – 1) A(–1, –1)

S = ( [5 + (-1) ]/2 , [(-1) + (-1)] / 2 ) = (2, -1)

using distance formula

P = (-1 , 3/2 ) ; Q= (2,4)

PQ = sqrt { (-1 -2)^2 + [(3/2) -4]^2 } =sqrt[61/4]

Q= (2,4) R=(5, (3/2) )

QR = sqrt{ (2-5)^2 + [4-(3/2)]^2 } = sqrt[61/4]

R=(5, (3/2) ) S =(2, -1)

RS = sqrt{(5-2)^2 + [(3/2) -(-1)]^2} = sqrt[61/4]

S =(2, -1) P = (-1 , 3/2 )

SP= sqrt{[2-(-1)]^2 + [(-1) - (3/2)]^2 } =sqrt[61/4]

similarly checking diagonals

P = (-1 , 3/2 ) R=(5, (3/2) )

PR = sqrt{16) = 4

S =(2, -1) Q= (2,4)

SQ =sqrt{25} =5

so diagonals are not equal

therefore all sides of the quadrilateral PQRS are equal, but the diagonals are not equal.

so PQRS is a rhombus.

exercise 7.3

Find the area of the triangle whose vertices are

(2, 3), (–1, 0), (2, – 4)

Area = (1/2) {2 [0-(-4)] +(-1)[(-4)-3] +2[3-0]}

=(1/2){8+7+6} = (21/2) sq units.

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ncert cbse 10th mathematics

co ordinate geometry chapter 7

exercise 7.4 optional exercise

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

solution

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

solution

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

exercise 7.3

Find the area of the triangle whose vertices are

(2, 3), (–1, 0), (2, – 4)

solution

Wednesday, May 5, 2021

The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of ∆ ADE and compare it with the area of ∆ ABC

exercise 7.4 optional exercise co ordinate geometry chapter 7 cbse ncert 10th mathematics

[AD/AB] =[AE/AC] =[1/4]

means

AB =4AD and AC=4AE

[ draw a line AB and plot D subject to AB =4AD ]

This shows that

AD / DB=1/3 so that D divides AB internally in the ratio 1:3

similarly E divides AC in the ratio 1:3 internally.

A(4, 6), B(1, 5) ratio =1:3

using section formula

D = ( [1*1 + 3*4 ]/[1+3] , [1*5+3*6 ]/[1+3] )

D = ( 13/4 , 23/4 )

A(4, 6), C(7, 2) ratio =1:3

E = ( [ 1*7 + 3*4 ]/ [1+3] , [ 1*2 +3*6]/[1+3] )

E = (19/4 , 5 )

A(4, 6), B(1, 5) and C(7, 2)

area of triangle ABC =(1/2)[4(5-2) +1(2-6)+7(6-5)]

=(1/2)[12+(-4)+7]=(15/2)

A(4, 6), D ( 13/4 , 23/4) , E = (19/4 , 5 )

area of triangle ADE =(1/2) [ 4( (23/4) -5 ) +(13/4)( 5 - 6 )+(19/4)( 6- (23/4)) ]

=(1/2) [ 4(3/4) +(13/4)(-1) +(19/4)(1/4) ]

=(1/2) [15/16] =[15/32]

[area of ADE] / [area of ABC] =[15/32] / [15/2] = 1/16

ratio = 1:16

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

(i)

D is the midpoint of BC

D = ( [6+1]/2 , [5+4]/2 )

D= (7/2 , 9/2)

(ii)

A = (4, 2),

D= (7/2 , 9/2)

ratio = 2:1

P = ( [ 2*(7/2) + 1*4 ] / [2+1] , [ 2(9/2) +1*2]/[2+1] )

P=( [11/3] , [11/3])

=================================================

ncert cbse 10th mathematics

co ordinate geometry chapter 7

exercise 7.4 optional exercise

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

solution

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1

solution

Monday, May 3, 2021

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

exercise 7.4 optional exercise co ordinate geometry chapter 7 cbse ncert 10th mathematics

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

let the given points be A(6, – 6), B(3, – 7) and C(3, 3).

let P(h,k) be the centre

using distance formula and the identities

PA = sqrt [ (h-6)^2 + (k+6)^2 ] = sqrt[ (h^2)+(k^2) -12h+12k+72]

PB=sqrt[ (h-3)^2 + (k+7)^2 ] = sqrt[ (h^2)+(k^2) -6h+14k+58]

PC=sqrt[ (h-3)^2 + (k-3)^2 ] = sqrt[ (h^2)+(k^2) -6h-6k+18]

Since P is the centre PA=PB=PC

PA=PB means on squaring

[ (h^2)+(k^2) -12h+12k+72] =[ (h^2)+(k^2) -6h+14k+58]

-6h -2k =(-14)--------------------(1)

similarly PA = PC gives

[ (h^2)+(k^2) -12h+12k+72] =[ (h^2)+(k^2) -6h-6k+18]

-6h +18k =(-54) ---------------------------(2)

solving (1) and (2)

-6h -2k =(-14)--------------------(1)

-6h +18k =(-54) ---------------------------(2)

--------------------------------------------------------------------------- subtracting

-20 k =(+40)

k= (40)/ (-20)

k = (-2)

substitute in -6h -2k =(-14)

-6h + 4 = (-14)

-6h = -14 -4

-6h = (-18)

h = (-18) /(-6)

h = 3

centre is P(h,k) =( 3 , -2 )

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

let the given vertices be A(-1,2) and C(3,2)

let B(h,k) be one of the other vertices

using distance formula

AB =sqrt [ (h+1)^2 +(k-2)^2 ] =sqrt[ (h^2) + (k^2) +2h-4k+5 ]

BC = sqrt[(h-3)^2 +(k-2)^2 ] =sqrt[ (h^2) + (k^2) -6h-4k+13 ]

AC = sqrt[(3+1)^2 + (2-2)^2] = 4

since all sides of a square are equal

AB= BC gives

[ (h^2) + (k^2) +2h-4k+5 ] = [ (h^2) + (k^2) -6h-4k+13 ]

8h = 8

h=1

now use pythagoras theorem in triangle ABC to solve for k

(AB)^2 + (BC)^2 = (AC)^2

[ (h^2) + (k^2) +2h-4k+5 ] + [ (h^2) + (k^2) -6h-4k+13 ] = 16

2 (h^2) +2(k^2) -4h -8k +2=0

apply h=1

2 +2(k^2) -4 -8k+2 = 0

2(k^2) -8k=0

2k(k-4) =0

k=0 or k=4

so other vertices are (1,0) and (1,4)

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ncert cbse 10th mathematics

co ordinate geometry chapter 7

exercise 7.4 optional exercise

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

solution

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution

Sunday, May 2, 2021

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

exercise 7.4 optional exercise co ordinate geometry chapter 7 cbse ncert 10th mathematics

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

let the ratio be k : 1

using section formula for internal division

point of intersection is

P [ (3k + 2) / (k+1) , (7k - 2) / (k+1) ]

This lies on 2x + y – 4 = 0

2 [ (3k + 2) / (k+1) ] + [(7k - 2) / (k+1) ] - 4 =0

multiplying by (k+1) and rearranging

6k+ 4 +7k-2 =4k+4

9k = 2

k = 2/9

ratio is 2 : 9 internally.

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

points are collinear means that

area of the triangle formed by the points = 0

(1/2) [x (2-0)+1(0-y )+7(y-2) ] =0

multiply off 2

2x -y +7y-14 =0

2x+6y -14 =0

dividing by 2

x +3y -7 =0 is the required relation

=================================================

ncert cbse 10th mathematics

co ordinate geometry chapter 7

exercise 7.4 optional exercise

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

solution

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution

chapter 5 arithmetic progressions

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term?

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m. Calculate the total volume of concrete required to build the terrace.

solution

chapter 5 arithmetic progressions, exercise 5.3

exercise 5.3

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

solution

19.

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on . In how many rows are the 200 logs placed and how many logs are in the top row?

solution

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . What is the total length of such a spiral made up of thirteen consecutive semicircles.

solution

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution

16. A sum of Rs.700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs.20 less than its preceding prize, find the value of each of the prizes.

solution

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs.300 for the third day, etc., the penalty for each succeeding day being Rs.50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

solution

14. Find the sum of the odd numbers between 0 and 50.

solution

13. Find the sum of the first 15 multiples of 8.

solution

12. Find the sum of the first 40 positive integers divisible by 6.

solution

11.If the sum of the first n terms of an AP is 4n –(n^2) , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

solution

10. Show that a1 , a2 , . . ., an , . . . form an AP where a n is defined as below :
an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

solution

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

solution

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

solution

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

solution

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

solution

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

solution

4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

solution

given a = 5, d = 3, an = 50, find n and Sn

solution

(ii) given a = 7, a13 = 35, find d and S13

solution

(iii) given a(12) = 37, d = 3, find a and S(12 )

solution

(iv) given a3 = 15, S(10) = 125, find d and a(10)

solution

(v) given d = 5, S9 = 75, find a and a9 .

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(vi) given a = 2, d = 8, Sn = 90, find n and an .

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vii) given a = 8, an = 62, Sn = 210, find n and d

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(vii) given an = 4, d = 2, Sn = –14, find n and a.

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ix) given a = 3, n = 8, S = 192, find d.

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(x) given L= 28, S = 144, and there are total 9 terms. Find a.

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find the sums given below :

7 + [10 +(1/2) ] +14 + ...+84

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(ii) 34 + 32 + 30 + . . . + 10

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(iii) –5 + (–8) + (–11) + . . . + (–230)

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Find the sum of the following APs:
2, 7, 12, . . ., to 10 terms.

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(ii) –37, –33, –29, . . ., to 12 terms.

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(iii) 0.6, 1.7, 2.8, . . ., to 100 terms

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(iv) (1/15) +(1/12) +(1/10) + .... 11terms

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Aruna saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs.20.75, find n.

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exercise 5.2

19. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs.200 each year. In which year did his income reach Rs.7000?

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18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

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17.Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

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16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

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15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . .

and 3, 10, 17, . . . equal?

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14. How many multiples of 4 lie between 10 and 250?

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