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.
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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.
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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.
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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)
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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]
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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
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4
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= –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]
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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
____________________________________________________________
…
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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
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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
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8
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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.
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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
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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
)
____________________________________________________________
…
_________________________________________________________________________________
= 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]
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