Skip to main content

ARITHMETIC PROGRESSIONS AND



ARITHMETIC PROGRESSIONS AND


CO-ORDINATE GEOMETRY


Time: 1 hour


Maximum Marks: 38


General instructions:


(i)  All the questions are compulsory.


(ii)  The question paper consists of 18 questions. Question number 1–10 are of 1 mark each, question


number 11–12 are of 2 marks each, question number 13–16 are of 3 marks each, question


number  17–18 are of 6 marks each.


(iii)  Internal choices have been provided in 2 marks, 3 marks and 6 marks questions. You have to


attempt one of the choices in such questions.


(iv)  Write down the serial number before attempting the question.


(v)  Use of calculators is not permitted.


1.  Determine whether the following sequence is an A.P. or not.


19, 32, 46, 59, ….


2.  Find the number of terms of the A.P whose first term is 9, common difference is 7 and the last term is


114.


3.  Find the mid point of the line segment joining the points (–3, 7) and (5, 6).


4.  Find the common difference and the next term of the A.P.


13, 9, 5, 1, –3, –7,…..


5.  Find the distance between the points (1, 2) and (2, 1).


6.  Find the ordinate of the point which divides the line segment joining the points (10, 6) and (3, 5) in the


ratio 2: 3, internally.


7.  Find the sum of the first 12 terms of an A.P. whose first term is 16 and common difference is –4.


8.  Find the 27


term of an A.P. whose first term is 7 and common difference is 15.


t h


9.  Find the area of a triangle whose vertices are (0, 0), (0, 7) and (6, 0).


10.  Find the distance between (a, 2a) and origin.


Time Tracker:


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 



_________________________________________________________________________________



11.  The distance between the points (2, 3) and (a, 2) is 5 units.  Find the value(s) of a.


12.  How many terms do the A.P 77, 80, 83, ……………167 have?


OR


The 30


term of an A.P is 45 and the 60


term of the A.P is –45. Find its first term and the common


th


t h


difference.


Time Tracker:


1 , –3) and (2, 0).


13.  Find a relation between x and y such that (x, y) is equidistant from ( 2


14.  How many terms of the sequence 18, 16, 14… should be taken so that their sum is zero?


OR


Mr. Sharma booked a room on each of the 15 floors of a hotel for his guests. The rent of a room on the


first floor is Rs. 1000 per day and for each next floor, the rent increases by Rs 300.  How much rent did


Mr. Sharma pay for the stay of his guests for 3 days in the hotel?


15.  In what ratio does the point (1, 1) divide the line segment joining the points (4, 0) and (–5, 3)?


16.  Show that the points A (–3, 4), B (1, 7) and C (–3, 7) are the vertices of a right angled triangle. Name the


hypotenuse of this triangle.


Time Tracker:


17.  Find the difference between the sum of the first 20 two digit natural numbers which are divisible by 2


and the sum of the first 12 two digit natural numbers which are divisible by 6.


OR


(i)  Find the sum of the first 23 multiples of 11.


(ii)  Find the sum of all the odd numbers between 0 and 28.


18.  Recognize the figure formed by the points (0, 3), (3, 0), (0, –3), (–3, 0)


Time Tracker:


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 




_________________________________________________________________________________



SOLUTIONS


1. Here, a


= 19, a


= 32, a


= 46


1


2


3


Now, a


– a


= 13 and a


– a


= 14


2


1


3


2


Since a


– a


a


– a


2


1


3


2


the given sequence is not an A.P.


[1]


2.  We have


a = 9,  d = 7, l = 114


We know l = a + (n–1) d


[½]


114 = 9 + (n – 1) 7


114 = 9 + 7n–7


112 = 7n


n = 16


[½]


Number of terms = 16


3.  Let P(x, y) is the mid point of the line segment joining the points A(–3, 7) and B(5, 6)


-


3+


5


x  =  2


2


=  2


= 1


[½]


7+


6


y  =  2


13


=  2


= 6.5


[½]


(1, 6.5) is the mid point of the line segment joining the points A(–3, 7) and B(5, 6).


4. Here, a


= 13, a


= 9


1


2


Now, d = a


– a


= 9 – 13


2


1


= – 4


[½]


and a


= a


+ d


7


6


= –7 + (–4)


= – 11


[½]


____________________________________________________________


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 




_________________________________________________________________________________



5.  The distance between any two points (x


, y


) and (x


, y


) is


1


1


2


2


-


+


( -


x


x


)


(


y


y


)


[½]


2


2


2


1


2


1


The distance between the points (1, 2) and (2, 1)


2


2


=


( -


1


-


2


)


+


(


2


1


)


= 1


1+


=  2   units


[½]


6.  Let P(x, y) divide the line segment joining the points A (10, 6) and B (3, 5) in the ratio 2 : 3, internally.


+


3


(


6


)


2


(


5


)


y   =  2


[½]


+


3


18+


10


=  5


28


=  5


= 5.6


[½]


7.  Here a = 16, d = –4, n = 12


n [2a + (n–1) d]


We know S


=  2


[½]


n


12 [2 × 16 + (12 – 1) (–4)]


= 2


= 6 [32 + 11 × (–4)]


= 6 [32 – 44] = 6 ×  –12


= –72


[½]


8.  Here, a = 7, d = 15


Now, t


= a + 26d


[½]


27


= 7 + 26  × 15


= 7 + 390


= 397


[½]


9.  Area of the triangle having vertices (0, 0), (0, 7), (6, 0) is


1 [0 (7 – 0) + 0 (0) + 6 (0 – 7)]


[½]


=  2


1 (– 42)


=  2


____________________________________________________________


4


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 




_________________________________________________________________________________



= –21


Now area cannot be negative.


Negative sign is ignored.


Area of triangle is 21 square units.


[½]


10.  Co–ordinates of origin are (0, 0).


Therefore required distance =


( -


a


-


0


)


+


(


2


a


0


)


[½]


2


2


=


a + =


4


a


5


a


2


2


2


=  5 a units


[½]


11.  Distance between (2, 3) and (a, 2) is 5 units.


( -


a


- = 5


2


)


+


(


2


3


)


[½]


2


2


Squaring both sides


(a–2)


+ 1 = 5


[½]


2


2


(a–2)


= 5


–1


2


2


(a–2)


= 24


2


a–2 =  ± 24


[½]


a–2 =  ± 2 6


a–2 =  ± 2 6  or a – 2 = –2 6


a = 2 + 2 6  or a = 2–2 6


Hence, required values of a are


2 + 2 6 and 2 – 2 6


[½]


12.  a = 77


d = 80 – 77 = 3


[½]


t


= 167


n


We know


t


= a + (n – 1)d


[½]


n


167 = 77 + (n – 1) 3


167 = 77 + 3n –3


167 = 74 + 3n


3n = 167 –74


93          = 31


n =  3


[1]


____________________________________________________________


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 




_________________________________________________________________________________



OR


t


= a + 29d = 45    ………..(i)


[½]


3  0


t


= a + 59d = –45    ………..(ii)


[½]


6  0


(ii) – (i)


30d = –90


90 = –3


d = – 30


[½]


Put value of d in equation (i)


a + 29 (–3) = 45


a = 45 + 87 = 132


[½]


1 , –3) and B(2, 0).


13.  Let P(x, y) be equidistant from A( 2


PA  =  PB


[½]


2


1


-  =


()


2


2


2


x +


+


(


y


3


)


x -


-


2


+


(


y


0


)


[½]


2


Squaring both sides


1  – x + y


x


+  4


+9 + 6y = x


+ 4 – 4x + y


[½]


2


2


2


2


1  = –4x + 4


6y  – x + 9 +  4


21  = 0


3x +6y +  4


Which is the required relation.


[½]


14.  Clearly, the given sequence is an A.P. with first term 18 and common difference –2.


[½]


Let the sum of n terms be 0.


Sn = 0


n [2a + (n – 1)d] = 0


2


[½]


n [2 × 18 + (n – 1)] = 0


2


n[18 – (n – 1) = 0


18n –n


+ n = 0


2


n


– 19n = 0


2


____________________________________________________________


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 






_________________________________________________________________________________


n(n–19) = 0


n = 0 or 19


But n cannot be 0


n = 19


[1]


OR


Let the rent of the room on the first floor be the first term of the A.P and the increase in the rent be the


common difference.


Now, rent of the room on the first floor = Rs 1000


Increase in the rent of room on each next floor = Rs 300


A. P is 1000, 1300, 1600……..


[½]


Since, there are 15 floors in the hotel.


There are 15 terms in this A.P.


[½]


n (2a + (n–1)d)


We know S


=   2


[½]


n


15 (2×1000 + (15–1) 300)


Total rent for 15 rooms =  2


15 (2000 + 14 × 300)


=   2


15 (2000 + 4200)


=   2


15  × 3100


=   2


=   Rs. 46500


[1]


Total rent for a stay of 3 days = 3 × 46500


= Rs. 139500


[½]


15.  Let (1, 1) divide the line segment joining the points (4, 0) and (–5, 3) internally in the ratio m


: m


.


1


2


Using the section formula,


we get


-


5


m


+


4


m


3


m


(1, 1) =


,


1


2


1


m


+


m


m


+


m


1


2


1


2


-  and 1 =


+


5


m


4


m


3


m


1 =


[1]


1


2


1


+


+


m


m


m


m


1


2


1


3


Now using the x coordinate


-


+


5


m


4


m


1 =


1


2


+


m


m


1


2


m


+ m


= –5m


+ 4m


[½]


1


2


1


2


 


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 


_________________________________________________________________________________



6m


– 3m


= 0


1


2


2m


– m


= 0


1


2


2m


= m


1


2


m  = 2


1


1


m


2


m


: m


= 1 : 2


[½]


1


2


To verify the ratio, let’s take the y–coordinate


3


m


i.e.


1


m


+


m


1


2


Dividing numerator and denominator by m


2


3


m


1


1


×


3


m


2


=


2


m


1


+


+


1


1


1


m


2


2


3


2


=


=1


3


2


The point (1, 1) divides the line segment joining the points (4, 0) and (–5, 3) in the ratio 1 : 2.


[1]


Alternate


Let point (1, 1) divide the line segment joining the points (4, 0) and (–5, 3) in the ratio  k:1


Using section formula, we get


-


5


k


+


4


3


k


(1, 1) =


,


+


+


k


1


k


1


- and 1


+


5


k


4


3


k


=


1 =  1


[1]


+


k


+


1


k


Now using the x coordinate


4


-


5


k


1 =  1


+


k


k+1 = 4 –5k


[½]


6k = 3


1


k = 2


[½]


The required ratio is k:1


1 :1


i.e.  2


____________________________________________________________


8


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 


_________________________________________________________________________________



or 1:2


The point (1, 1) divides the line segment joining the points (4, 0) and (–5, 3) in the ratio 1:2


[1]


16.  The given points are


A (–3, 4), B (1, 7) and C(–3, 7)


Now, (AB)


= (


1 -


-


(


-


3


))


+


(


7


4


)


2


2


2


= (1 + 3)


+ 3


2


2


= 16 + 9


= 25


2


2


(BC)


=


( -


-


3


-


1


)


+


(


7


7


)


2


=  4


+  0


2


2


= 16


-


-


-


+


2


2


(CA)


=


( -


3


(


3


))


(


4


7


)


2


= (–3 + 3)


+ (–3


)


2


2


= 0 + 9


= 9


[1½]


Now, 16 + 9 = 25


(BC)


+ (CA)


= AB


[½]


2


2


2


By the converse of Pythagoras theorem


ABC is a right angled triangle and hypotenuse is AB.


[1]


17.  First two digit natural number divisible by 2 is 10


Second two digit natural number divisible by 2 is 12


Common difference = 12 – 10 = 2


This forms an A.P with first term 10 and common difference 2


Sum of the first 20 two digit natural numbers divisible by 2


20 [2 × 10 + (20 –1) 2]


=  2


= 10 [20 + 38]


=  10  × 58


= 580    …..(i)


[2½]


First two digit natural number divisible by 6 is 12


Second two digit natural number divisible by 6 is 18


Common difference = 18 – 12 = 6


These numbers form an A.P with first term 12 and common difference 6.


____________________________________________________________


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 


_________________________________________________________________________________



Sum of the first 12 two digit natural numbers divisible by 6


12


=


[2 × 12 + (12 –1) 6]


2


= 6 [24 + 11 × 6]


=  6  × [24 + 66]


=  6  × 90


= 540   …..(ii)


[2½]


Subtracting (ii) from (i), we get


580 – 540 = 40


or subtracting (i) from (ii), we get


540 –580 = –40


Required difference is 40 or – 40


[1]


OR


(i)  The first multiple of 11 is 11


The second multiple of 11 is 22


The third multiple of 11 is 33


Thus the multiples of 11 form an A.P. with the first term 11 and common difference 11.


i.e. 11, 22, 33 …..


[1]


Now, we want the sum of the first 23 multiples of 11


i.e. we want to find the sum of the first 23 terms of this A.P.


[½]


n [2a + (n–1)d]


Now, S


=  2


[½]


n


Here, a = 11, d = 11 and n = 23


Sum of the first 23 multiple of 11


23 [2× 11 +


= 2


(23 – 1)11]


23 [22 + 242]


= 2


23 ×  264


= 2


= 3036


[1]


(ii)  The first odd number is 1


The second odd number is 3 and the last odd number between 0 and 28 is 27


[1]


The odd numbers between 0 and 28 form an A.P. with first term 1, common difference 2 and


the last term 27


____________________________________________________________


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 


_________________________________________________________________________________



a = 1, d = 2 and a


= 27 (or  l = 27)


n


We know


a


= a + (n –1)d


[½]


n


27 = 1 + (n –1) 2


26 = n – 1


2


n – 1 = 13


n = 14


[½]


n (a +l )


Also, S


=  2


[½]


n


14 (1 + 27)


=  2


= 196


[½]


18.  Let the given points be A(0,3), B(3, 0), C(0, –3), D(–3, 0)


Now,


2


2


|AB| =


( -


3


-


0


)


+


(


0


3


)


=  9


9+  = 18


=3 2  units


[1]


|BC| =


( -


0


-


3


)


+


(


-


3


0


)


2


2


=  9


9+  = 18


=3 2  units


[1]


|CD| =


-


-


+


-


2


2


( -


3


0


)


(


0


(


3


))


= 9


9+  = 18


=3 2 units


[1]


-


-


+


2


2


|AD| =


( -


3


0


)


(


0


3


)


=  9


9+  = 18


=3 2  units


[1]


|AB| = |BC| = |CD|=|AD|


Now,


-


+


-


2


2


|AC| =


( -


0


0


)


(


3


3


)


____________________________________________________________


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 




_________________________________________________________________________________


=  36


= 6 units


2


2


|BD| =


( -


-


3


-


3


)


+


(


0


0


)


=  36


= 6 units


Thus, the figure formed by given points has all four sides equal and both the diagonals are equal to each


other as well.


The figure formed by (0,3), (3,0) (0,–3) and (–3,0) is a square.


[2]


Alternative method


We have to prove that


AB = BC = CD = AD = 3 2  units


[4]


and any one diagonal AC = 6 units


AB


+ BC


= (3 2)


+ (3 2)


2


2


2


2


= 18 + 18


= 36   = 6


2


AB


+ BC


= AC


2


2


2


By the converse of Pythagoras theorem


B = 90°


We know a quadrilateral with all the four sides equal and one angle equal to 90° is a square.


[2]


____________________________________________________________


-------------------------------------------------------------------------------------------------------
www.cbsekey.com
Other Educational Portals
English | cbse | Science | Mathematics


 



Comments

Popular posts from this blog

The Missing Mail | Class IX - Interact in English

NCERT / CBSE Literature Reader for English Course (Communicative) Important Exercise Questions Q.3: (a) Why is Ramanujam worried about getting his daughter married? Give four reasons. (b) How does the postman console and guide Ramanujam and his family during each of the instances you have listed in 3 (a)? Ans 3(a): Ramanujam is worried as he could not find a suitable match to marry his daughter off which was getting delayed because of different reasons. The four causes of his worriedness are - (i) Sometimes horoscopes did not match, (ii) Sometimes the girl’s appearance were not approved, (iii) At times there were problems of too much dowry and other financial matters, (iv) The season was closing with only three more auspicious dates left, whereas, he was not able to finalise any alliance by that time. Ans 3(b): First instance - When Ramanujam said that horoscopes did not agree Thanappa consoled and guided him by saying that he should not utter inauspicious words and when the God wills

The Tribute - Multiple Choice Questions

Question (1):   What does the word 'petite' mean? 1. strong 2. small 3. neat 4. clean Ans:   2 Question (2):   What was almost a routine affair for Babuli during his student days? 1. working in the city 2. writing to his elder brother 3. going home in the rickety bus 4. reading big books Ans:   3 Question (3):   What was written in the elder brother's letter? 1. That they were missing him in the village 2. That there was to be a partition 3. That there was a festival 4. That he would get money Ans:   2 Question (4):   How many days were left for the partition? 1. A week 2. A month 3. A day 4. Three days Ans:   1 Question (5):   How much money was expected for Babuli's land? 1. Seven thousand 2. Twenty thousand 3. Ten thousand 4. Forty thousand Ans:   2 Question (6):   What makes Babuli think of the butcher? 1. Hunger 2. Wife's heartlessness 3. The innocent lamb 4. The partition A

ENGLISH (Communicative) Sample Question Paper 5

Sample Paper – 2009 Class – XSubject – ENGLISH (Communicative) General instructions: The paper consist of FOUR sections: SECTION A (READING) - 20 Marks SECTION B (WRITING) - 30 Marks SECTION C (GRAMMAR) - 20 Marks SECTION D (LITERATURE) - 30 Marks Attempt all the questions. Do not write anything on the question paper. All the answers must be correctly numbered as in the question paper. And written in the answer sheets provided to you. Attempt all questions in each section before going on to the next section. Read each question carefully and follow the instructions. Strictly adhere to the word limit given with each question. Marks will be deducted for exceeding the word limit. SECTION A (READING) – 20 MARKS A1. Read the following passage and answer the following questions: [12] THE TUITION TRAP 1. Given the general awareness of the woeful condition of our State sch