Loading....
Coupon Accepted Successfully!

 

Question-1

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is `15 for the first km and `8 for each additional km.
(ii) The amount of air present in a cylinder when a vacum pump removes of the air remaining in the cylinder at a time.

(iii)The cost of digging a well after every metre of digging, when it costs `150 for the first metre and rises by `50 for each subsequent metre.
(iv) The amount of money in the account every year, when `10000 is deposited at compound interest at 8% per annum.

Solution:
(i) The taxi fare after each km when the fare is `15 for the first km and `8 for each additional km.
The taxi fare will be 15, 15 + 8, 15 + 16, 15 + 24, 15 + 32, 15 + 40 etc.
(i.e.,) 15, 23, 31, 39, 47, 55 etc.
This is an AP as every succeeding term is obtained by adding 8 in its preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.
Let amount of air present in the cylinder be "a".
The vacuum removes of air remaining in the cylinder.
The air present in the cylinder will be a, a - = , -==,
a, , , …
This is not an AP because the difference between second term and first term is not the same as the difference between the third term and the second term.

(iii) The cost of digging a well after every metre of digging, when it costs `150 for the first metre and rises by `50 for each subsequent metre.
The cost of digging the well will be `150, `(150 + 50), `(150 + 100), `(150 + 150) etc.
(i.e.,) `150, `200, `250, `300, `350 etc
This is an AP as every succeeding term is obtained by adding `50 in its preceding term.

(iv) The amount of money in the account every year, when `10000 is deposited at compound interest at 8% per annum.
Initial amount of money = `10,000
Compound Interest = 8% p.a
Amount every year increase by 8% Compound Interest p.a.
In the second year amount increases to 10,000
In the 3rd year amount increases to 10,000
In the 4th year amount increases to 10,000
This is not an AP as the difference is not the same in this problem.

Question-2

Write first four terms of the A.P, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = -3 (iv) a = -1, d =
(v) a = -1.25, d = -0.25

Solution:
(i) a = 10, d = 10
Let a1, a2, a3, a4 be the first four terms of the A.P
Here a = a1 = 10
a2 = a1 + d = 10 + 10 = 20
a3 = a2 + d = 20 + 10 = 30
a4 = a3 + d = 30 + 10 = 40
The first four terms of the A.P are 10, 20, 30, 40.
 
(ii) a = -2, d = 0
Let a1, a2, a3, a4 be the first four terms of the A.P
Here a = a1 = -2
a2 = a1 + d = -2 + 0 = -2
a3 = a2 + d = -2 + 0 = -2
a4 = a3 + d = -2 + 0 = -2
Thus first four terms of the A.P are -2, -2, -2, -2

(iii) a = 4, d = -3
Let a1, a2, a3, a4 be the first four terms of the A.P
Here a = a1 = 4
a2 = a1 + d = 4 + (-3) = 1
a3 = a2 + d = 1 + (-3) = -2
a4 = a3 + d = -2 + (-3) = -5
The first 4 terms of the A.P are 4, 1, -2, -5

(iv) a = -1, d =
Let a1, a2, a3, a4 be the first four terms of the A.P
Here a = a1 = -1
a2 = a1 + d = -1 + = -
a3 = a2 + d = - + = 0
a4 = a3 + d = 0 + =
The first 4 terms of the A.P are –1, -, 0, .

(v) a = -1.25, d = -0.25
Let a1, a2, a3, a4 be the first four terms of the A.P
Here a = a1 = -1.25
a2 = a1 + d = -1.25 + (-0.25) = -1.50
a3 = a2 + d = -1.50 + (-0.25) = -1.75
a4 = a3 + d = -1.75 + (-0.25) = -2.0
The first 4 terms of the A.P are -1.25, -1.50, -1.75, -2.0.

Question-3

For the following A.Ps, write the first term and the common difference:
(i) 3, 1, -1, -3, … (ii) –5, -1, 3, 7, … (iii) … (iv) 0.6, 1.7, 2.8, 3.9, …

Solution:
(i) The first term is 3 and common difference = a2 – a1= 1 - 3 = -2.
(ii) The first term is -5 and common difference = a2 – a1= (-1) – (-5) = 4. 
(iii) The first term is and common difference = a2 – a1= - = .
(iv) The first term is 0.6 and common difference = a2 – a1= (1.7) – (0.6) = 1.1.

Question-4

Which of the following are A.Ps? If they form an A.P, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, … (ii) 2,
, … (iii) –1.2, -3.2, -5.2, -7.2, …
(iv) –10, -6, -2, 2, … (v) 3, 3 +, 3 + 2, 3 + 3, … 

Solution:
(i) 2, 4, 8, 16, …
Here a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
Since ak+1 - ak is not the same throughout, it is not an A.P.

(ii) 2, , …
Here
a2 – a1 = – 2=
a3 – a2 = 3 – =
a4 – a3 = – 3 =
As ak+1 - ak is the same throughout, it is an A.P with first term = 2 and common difference =
The next three terms are 4, 4 + = , + = = 5.

(iii) –1.2, -3.2, -5.2, -7.2, …
Here
a2 – a1 = -3.2 – (-1.2) = -3.2 + 1.2 = -2
a3 – a2 = -5.2 – (-3.2)= -2
a4 – a3 = -7.2 – (-5.2)= -2
Since ak+1 – ak is the same throughout it is an A.P with first term (a) = -1. 2 and common difference, (d) = -2
Thus the next three terms are
a5 = a4 + d = –7.2 – 2 = -9.2
a6 = a5 + d = -9.2 – 2 = -11.2
a7 = a6 + d = - 11.2 – 2 = -13.2

(iv) –10, -6, -2, 2, …
Here
a2 – a1 = -6 – (-10)= 4
a3 – a2 = -2 – (-6) = 4
a4 – a3 = 2 – (-2)= 4
As ak+1 – ak is the same throughout it is an A.P with 1st term as -10 and common difference as 4.
Thus the next three terms are
a5 = a4 + d = 2 + 4 = 6
a6 = a5 + d = 6 + 4 = 10
a7 = a6 + d = 10 + 4 = 14

(v) 3, 3 +, 3 + 2, 3 + 3, …
Here a2 – a1 = 3 +  - 3 =
a3 – a2 = 3 + 2- (3 +) =
a4 – a3 = 3 + 3- (3 + 2) =
In this problem, ak+1 –ak is the same throughout thus the given list is an A.P with 1st term as 3 and common difference (d) as .
Thus the next three terms are
a5 = a4 + d = (3 + 3) += 3 + 4
a6 = a5 + d = (3 + 4) + = 3 + 5
a7 = a6 + d = (3 + 5) += 3 + 6

Question-5

Which of the following are A.Ps? If they form an A.P, find the common difference d and write three more terms.
(i) 0.2, 0.22, 0.222, 0.222, … (ii) 0, -4, -8, -12, … (iii), … (iv) 1, 3, 9, 27, … (v) a, 2a, 3a, 4a, …

Solution:
(i) 0.2, 0.22, 0.222, 0.2222, …
Here,
a2 – a1 = 0.22 – 0.2 = 0.02
a3 – a2 = 0.222 – 0.22 = 0.002
a4 – a3 = 0.2222 – 0.222 = 0.0002
Hence ak+1 – ak is not the same throughout, hence it is not an A.P.

(ii) 0, -4, -8, -12
Here
a2 – a1 = -4 – 0 = -4
a3 – a2 = -8 – (-4) = -4
a4 – a3 = -12 – (-8) = -4
Here ak+1 – ak is the same throughout hence it is an A.P with 1st term = 0
and common difference = -4
Thus the next three terms are
a5 = a4 + d = -12 + (-4) = -16
a6 = a5 + d = -16 + (-4) = -20
a7 = a6 + d = -20 + (-4) = -24

(iii) , …
Here
a2 – a1 = - = 0
a3 – a2 = = 0
a4 – a3 = = 0
 
Here ak+1 – ak = 0 is the same throughout hence the list is an A.P with 1st term = and common difference d = 0.
Thus the next three terms are
a5 = a4 + d = + 0 =
a6 = a5 + d = + 0 =
a7 = a6 + d = + 0 = .

(iv) 1, 3, 9, 27, …
Here the 1st term (a) = 1
a2 – a1 = 3 – 1 = 2
a3 – a2 = 9 - 3 = 6
a4 – a3 = 27 – 9 = 18
Since ak+1 – ak is not the same throughout this is not an A.P.

(v) a, 2a, 3a, 4a
Here
a2 – a1 = 2a - a = a
a3 – a1 = 3a – 2a = a
a4 – a3 = 4a – 3a = a
Since ak+1 – a1 is the same throughout this is a A.P with first term = a and common difference = a
Thus the next three terms are
a5 = a4 + d = 4a + a = 5a
a6 = a5 + d = 5a + a = 6a
a7 = a6 + d = 6a + a = 7a

Question-6

Which of the following are A.Ps? If they form an A.P, find the common difference d and write three more terms.
(i) a, a2, a3, a4, … (ii) ,,, , …
(iii) , … (iv) 12, 32, 52, 72, … (v) 12, 52, 72, 73, …

Solution:
(i) a, a2, a3, a4, …
Here
a2 – a1 = a2 - a = a(a – 1)
a3 – a2 = a3 – a2 = a2(a – 1)

a4 – a3 = a4 – a3 = a3(a – 1)
Since ak+1 – ak is not the same throughout this is not an A.P.
  
(ii) ,,,, …
Here
a2 – a1 = - = 2 - =
a3 – a2 = - = 3 - 2 =
a4 – a3 = - = 4 - 3 =
Since ak+1 – ak is the same throughout it is an A.P with the first term as and common difference d =.
Thus the next three terms are
a5 = a4 + d = + = 4 + = 5
a6 = a5 + d = 5 + = 6
a7 = a6 + d = 6 + = 7.

(iii)
Here
a2 – a1 = = ( - 1)
a3 – a2 = = 3 -= (- )
a4 – a3 = = 2 - 3 = (2 - ).
Here ak+1 – ak is not the same throughout hence the list is not an A.P.

(iv) 12, 32, 52, 72
Here
a2 – a1 = 32 – 12 = 9 – 1 = 8
a3 – a2 = 52 – 32 = 25 – 9 = 16
a4 – a3 = 72 – 52 = 49 – 25 = 24
Since ak+1 – ak is not the same throughout hence the list not an A.P.

(v) 12, 52, 72, 73, …
Here
a2 – a1 = 52 – 12 = 25 – 1 = 24
a3 – a2 = 72 – 52 = 49 – 25 = 24
a4 – a3 = 73 – 72 = 73 – 49 = 24
Since ak+1 – ak = 24 is the same throughout this list is an A.P with first term = 12 and common difference = 24.
Thus the next three terms are
a5 = a4 + d = 73 + 24 = 97
a6 = a5 + d = 97 + 24 = 121
a7 = a6 + d = 121 + 24 = 145.

Question-7

Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P:
 

a

d

n

an

(i)

(ii)

(iii)

(iv)

(v)

7

-18

….

-18.9

3.5

3

….

-3

2.5

0

8

10

18

105

...

0

-5

3.6

...

 
Solution:
(i) a = 7, d = 3, n = 8, an = ?
an = a + (n – 1) × d = 7 + (8 - 1) × 3
    = 7 + 7 × 3 = 7 + 21 = 28. 

(ii) a = -18
n = 10, an = 0, d = ?
an = a + (n – 1) × d
0 = -18 + (10 – 1) × d
0 = -18 + 9 × d
0 = -18 + 9d
9d = 18
d = 2.

(iii) a = ?, d = -3, n = 18, an = -5
an = a + (n – 1) × d
-5 = a + (18 – 1) × (-3)
-5 = a + 17 × (-3)
-5 = a - 51
a = 51 – 5 = 46. 

(iv) a = -18.9
d = 2.5, an = 3.6, n = ?
an = a + (n – 1) × d
3.6 = -18.9 +(n – 1) × 2.5
(n – 1) × 2.5 = 3.6 + 18.9
(n – 1) = = ( To remove the decimal multiply and divide by 10)
 n –1 = 9
n = 10.

(v) a = 3.5, d = 0, n = 105
an = ?
an = a + (n – 1) × d
    = 3.5 +(105 – 1) × 0
an = 3.5.

Question-8

Choose the correct choice in the following and justify:
(i) 30th term of the A.P: 10, 7, 4, …, is
(A) 97 (B) 77 (C) –77 (D) – 87
(ii) 11th term of the A.P: -3,  , 2, …, is
(A) 28 (B)22 (C) –38 (D) -48

Solution:
(i) 30th term of the A.P: 10, 7, 4, …, is
nth term = a + (n – 1)d, where n = 30
30th term = 10 + (30 – 1) × (-3)
              = 10 + (29 × - 3)
              = 10 – 87 = - 77
(C) is the correct answer.

(ii) 11th term of the A.P: -3,  , 2, …, is
Here n = 11, 
a = -3
d = – (-3)
  = = = 2
an = a + (n – 1)d
an = -3+ (n – 1) × (2)
    = -3 +(11 – 1) × = -3 + 10 ×
    =– 3 + 25 = 22
a13 = 22.
(B) is the correct answer.

Question-9

In the following A.Ps, find the missing terms in the boxes:
(i) 2, ,26 (ii) , 13, , 3 (iii) 5, , , 9

(iv) -4,, , , , 6    (v) , 38, , , , -22

Solution:
(i) Let x be the missing term
x - 2 = 26 - x
2x = 28
x = 14
Missing term = 14

(ii) , 13, , 3
Let missing term in 1st box be x, missing term in the 2nd box be y,
13 - x = y - 13 = 3 - y
Now considering the last two,
y - 13 = 3 - y
2y = 16y = 8.
The value of y = 8.
Here 13 - x = y - 13
13 - x = 8 - 13
x = 26 - 8= 18.

(iii) 5, , , 9
Here a = 5, a4 = 9n = 4d = ?
an = a + (n- 1) × d

9 = 5 + (4 - 1) × d
= 5 + 3d
3d = - 5 = =
d = =
If a1 = 5
a2 = a1 + d = 5 + == = 6
a3 = 6 + = 8

(iv) -4, , , , , 6
an = 6
a = -4
d = ?

an = a + (n -1) × d

6 = -4 + (6 -1) × d
6 = -4 + 5d
5d = 6 + 4 = 10

d = = 2
-4, -2, 0, 2, 4, 6 is the A.P series.

(v) , 38, , , , -22
a2 = 38, n = 6, a6 = -22
an = a + (n - 1) × d if n = 6,
-22 = a + 5d ……. (1)
if n = 2,
38 = a + d ………. (2)

- (2)
⇒ 4d = -60
d = = -15.
Thus a2 = a + d = a + (-15)
38 = a - 15
a = 38 + 15 = 53.
a3 = 38 + (-15) = 23
a4 = 23 + (-15) = 8
a5 = 8 + (-15) = -7
53, 38, 23, 8, -7, -22 is the A.P series

Question-10

Which term of the A.P: 3,8, 13, 18, …,is 78

Solution:
Let 78 be the nth term
Here a = 3
d = 5, an = 78
an = a + (n –1) × d
78 = 3 + (n – 1) × d
78 – 3 = (n –1) × 5
75 = (n – 1) 5
= (n – 1)
n – 1 = 15
n = 16
78 is the 16th term.

Question-11

Find the number of terms in each of the following A.Ps:
(i) 7, 13, 19, …, 205 (ii) 18, 15
, 13, …, 47

Solution:
(i) Let the number of terms be n
Here a = 7, d = 6, an = 205
an = a + (n – 1) d
205 = 7 + (n – 1) × 6

(n – 1)6 = 205 – 7
n – 1 =

n = 33 +1 = 34
Number of terms = 34.

(ii) 18, 15, 13, …, 47
Here a = 18
d = 18 – 15 = -2 
an = -47
Let the number of terms be n
an = a + (n – 1) d
-47 = 18 + (n –1) (-2 )
-47 – 18 = (n – 1) ()
-65 × = (n – 1)
26 + 1 = n
Number of terms = 27

Question-12

Check whether –150 is a term of the A.P: 11, 8, 5, 2 ………

Solution:
Here a = 11, d = -3
If an = -150 be the nth term of the A.P
an = a + (n – 1) × d
-150 = 11 + (n – 1) × (-3)
-150 - 11 = (n – 1) × (-3)
= (n – 1)
n = = 53 + 1
which is not a whole number
-150 is not a term of the A.P.

Question-13

Find the 31st term of an A.P whose 11th term is 38 and the 16th term is 73.

Solution:
11th term is 38 and
16th term is 73
a1138 = a + 10d …………….(1)
a1673 = a +15d ………………..(2)
a + 10d = 38
a + 15d = 73
5d = 35
d = 7
sub d = 7 in (1)
a + 10 d = 38
a + 10(7) = 38
a = -70 + 38
  = - 32
Thus 31st term = a + (31 – 1) × d
                     = 32 + 30 × 7
                     = – 32 + 210
                     = 178.

Question-14

An A.P consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:
3rd term is a3 = a + (3 – 1)d = 12 ………………….. (1)
50th term is a50 = a + (50 – 1)d = 106 ………………….. (2)
(1) a + 2d = 12 ……………….. (3)
(2) a + 49d = 106 ……………….. (4)
-----------------
(3) – (4) Þ 47d = 94
d = 2
sub d = 2 in (3)
a + 2(2) = 12
a = 12 – 4 = 8
29th term of an A.P is a + (29 – 1)d ….(5)
substituting for a = 8 and d = 2 in (5),
a + 28d = 8 + 28 × 2
            = 8 + 56
            = 64.

Question-15

If the 3rd and the 9th terms of an A.P are 4 and –8 respectively, which term of this A.P is zero?

Solution:
3rd term of the A.P is a + (3 –1)d = 4
(i.e.,) a + 2d = 4 …………. (1)
9th term of the A.P is a + (9 – 1)d = -8
a + 8d = -8 …………………….(2)
a + 2d = 4…….(1)
a + 8d = -8…..(2)

(1) – (2)
 -6d = 12
d = -2
substituting d = -2 in (1) ⇒ a – 4 = 4
a = 8
Let the nth term be 0 then a + (n – 1) d = 0
(i.e.,) 8 + (n – 1) (-2) = 0
-2n + 2 = -8
-2n = -8 –2= -10
⇒ n = 5
Thus n = 5. Hence the 5th term is zero.

Question-16

The 17th term of an A.P exceeds its 10th term by 7. Find the common difference.

Solution:
17th term of the A.P = a +(17 – 1)d
                           = a + 16 d ………………………….. (1)
10th term of the A.P = a + (10 – 1)d = a + 9d ------------- (2)
Given (1) exceeds (2) by 7
[a + 16d] – [a + 9d] = 7
16d – 9d = 7
7d = 7
d = 1

Question-17

Which term of the A.P: 3, 15, 27, 39, … will be 132 more than its 54th term?

Solution:
54th term of the A.P is a + (54 - 1)d
Here a = 3
d = a2 – a1 = 15 – 3 = 12
54th term = 3 + (54 – 1) × 12
                  = 3 + 53 × 12
                  = 3 + 636 = 639
132 more than 54th term = 639 + 132 = 771
To find out which term of the A.P is 771, let us assume it is the nth term
a + (n – 1) × d = 771
substituting for a = 3 and d = 12
3 + (n – 1) × 12 = 771
(n – 1) × 12 = 771 - 3
(n – 1) = = 64
n = 65

Hence the 65th term is 132 more than its 54th term.

Question-18

Two A.Ps have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:
Let a1 and a2 be the 1st term of the 2 A.Ps and "d" be the common difference.
100th term of 1st A.P = a1 + (100 – 1)d ………………………. (1)
100th term of 2nd A.P = a2 + (100 – 1)d ………………………. (2)
Difference between (1) and (2) = 100
a1 + 99d – (a2 + 99d) = 100
(i.e.,) a1 – a2 = 100 …………….(3)
Difference between then 1000th terms = (a1 + 999d) – (a2 + 999d)
= a1 – a2 = 100 (from (3))
Thus the difference between their 1000th terms = 100.

Question-19

How many three-digit numbers are divisible by 7?

Solution:
The list of 3 digit numbers divisible by 7 are 105, 112, 119, …, 994.

This is an A.P. with a = 105, d = 7

 

an = 994

994 = a+(n – 1) × d

         = 105+(n – 1) × 7

7(n – 1) = 994 – 105

            = 889

n – 1 = = 127

n = 128

Thus the number of three-digit numbers divisible by 7 are 128.


Question-20

How many multiples of 4 lie between 10 and 250?

Solution:
Multiples of 4 between 10 and 250 are 12, 16, 20, …, 248.
Here
a = 12
d = 4
an = 248
248 = 12 + (n – 1) × 4
(n – 1) × 4 = 248 - 12
(n – 1) =
n = 59 + 1
   = 60
Number of multiples of 4 between 10 and 250 = 60.

Question-21

For what value of n, are the nth terms of two A.Ps: 63, 65, 67, …, and 3, 10, 17, …, equal?

Solution:
nth term of the 1st A.P = a + (n –1) × d
Here a = 63, d = 2
 63 + (n - 1) × 2 …………….(1)
nth term of the 2nd AP = a + (n –1) × d
Here a = 3, d = 7
 3 + (n – 1) × 7 ……………….(2)
63 + ( n – 1) × 2 = 3 + (n – 1) × 7 (Given)
63 – 3 = 7n – 7 – 2n + 2
60 + 5 = 5n
5n = 65
n = = 13.
13th term of the 2 A.Ps are equal.

Question-22

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

Solution:
3rd term = a + (3 - 1) × d
            = a + 2d = 16 ……………. (1)
7th term = a + 6d
5th term = a + 4d
Given (a + 6d) – (a + 4d) = 12
2d = 12 d = 6
substitute d = 6 in (1)
a + 2(6) = 16
a + 12 = 16
a = 16 – 12= 4.
a = 4, d = 6
∴ The A.P is 4, 10, 16, …

Question-23

Find the 20th term from the last term of the A.P: 3, 8, 13, …, 253.

Solution:
Here
a = 3, d = 5
Last term of the A.P is 253
Let 253 be the nth term of A.P
nth term = a + (n – 1) × d
              = 3 + 5(n – 1)
5n – 5 = 250
5n = 255
n = 51
253 is the 51st term.
20th term from the 51st term is 32nd term
32nd term = a + (32 – 1)d
                 = 3 + 31 × 5
                 = 3 +155
                 = 158.

Question-24

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

Solution:
4th term = a + 3d
8th term = a + 7d
Sum of the 4th term and 8th term = a + 3d + a + 7d = 24
2d + 10d = 24
a + 5d = 12 …………………………..(1)
Sum of 6th term and 10th term = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ……………………..………(2)
a + 5d = 12 ……. (1)
a + 7d = 22 …….. (2)
--------------
subtracting (1) – (2) ⇒ 2d = 10
d = 5
substituting d = 5 in (1)
a + 5d = 12
a + 5(5) = 12
a = 12 – 25 = -13.
∴ The 1st 3 terms are –13, -13 + 5, -13 + 10
(i.e.,) -13, -8, -3

Question-25

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

Solution:
Annual salary = `5000
Increment = `200
Total salary after n years = `7000
If the A.P. was of the form 5000, 5200, 5400, …
When 1st term a = 5000, d = 200
nth term = a + (n – 1) d = 7000
              = 5000 +(n – 1) × 200 = 7000
200(n – 1) = 7000 – 5000 = 2000
∴ (n – 1) =
n = 10 + 1
His income reached `7000 in the 11th year

Question-26

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

Solution:
Here

a = 5

d = 1.75

In the nth week saving’s = 5 + 1.75(n – 1) = 20.75

 

1.75(n – 1) = 20.75 – 5

n - 1 =

n -1 = 9

n = 10.


Question-27

In an A.P:
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5 , S9 = 75, find a and a9.

Solution:
(i) Given a = 5, d = 3, an = 50, find n and Sn.
a = 5, d = 3, an = 50
n = = = = 16
Sn =
S16 =  = 8 × 55 = 440.
Thus n = 16 and S16 = 440.

(ii) Given a = 7, a13 = 35, find d and S13.
a = 7, a13 = 35 find d and S13
a = 7, l = 35, n = 13
We know that n =
13 =
13 – 1 =
∴ d = =
Sn =
S13 =
      = = 273.
Thus d = and S13 = 273.

(iii) Given a12 = 37, d = 3, find a and S12.
a12 = l = 37, d = 3,
n = 12, a = ?

n =
12 =

37 – a = 11 × 3
-a = 33 - 37= -4
a = 4
Sn =
S12 =
      = = 246.
Thus a = 4 and S12 = 246.

(iv) Given a3 = 15, S10 = 125, find d and a10.
a3 = 15, S10 = 125
a +2d = 15 …………………….. (1)
S10 = 125
(2a + 9d) = 125
⇒ 5 (2a + 9d ) = 125
2a + 9d = = 25
2a + 9d = 25……………… (2)
(1) ×2 ⇒ 2a + 4d = 30 …..(3)
Subtracting (2) from (3) weget,
 -5d = 5
 
∴ d = -1
Substituting d = -1 in (1), we get,
a + 2d = 15
a + 2(-1) = 15
a = 17
a10 = a + 9d = 17 + 9(-1)
      = 17 – 9 = 8.
Thus d = -1 and a10 = 8.

(v) Given d = 5 , S9 = 75, find a and a9.
S9 = 75 , n = 9
an = a +(n – 1)5
a9 = a + 8 × 5 a9 = 40 + a …………….. (1)
Sn =
S9 = = 75
2a + 40 = 75 × ………………(2)
2a = – 40
2a = =
a = =
Substituting a = in (1), we get,
a9 = 40 + = =
Thus a = and a9 = .

Question-28

In an A.P:
(i) Given a = 2 , d = 8, Sn = 90, find n and an.
(ii) Given a = 8, an = 62, Sn = 210, find n and d.
(iii) Given an = 4, d = 2, Sn = -14, find n and a.
(iv) Given a = 3, n = 8, S = 192, find d.
(v) Given l = 28, S = 144, and there are total 9 items. Find a.  

Solution:
(i) Given a = 2 , d = 8, Sn = 90, find n and an.
a = 2, d = 8
Sn = 90, find n and an
Sn =
    =
    =
But Sn = 90,
=> n[8n – 4] = 90 × 2
8n2 – 4n – 180 = 0
Dividing throughout by 4, we get
2n2 – n – 45 = 0
2n2 –10n + 9n – 45 = 0
2n(n - 5) + 9(n - 5) = 0
(2n + 9)(n – 5) = 0
n = 5 (or) n =
Number of terms could be positive only, thus
n = 5 , and n =   not possible
an = [a + (n – 1) × d]
    an    = [2 + (5 – 1) × 8] = 2 + 32 = 34.
Thus n = 5 and an= 34.

(ii) Given a = 8, an = 62, Sn = 210, find n and d.
Here, a = 8, an = 62, Sn = 210
Now we have to find n and d,
an = a + (n – 1) × d
    = 8 + (n – 1) × d
(n – 1) × d = 62 – 8 = 54 ……………………..(1)
Sn =
210 =
420 = ……………………. (2)
Substituting for (n – 1) × d = 54 from (1) in (2)
420 = n[16 +54]
∴ n = = 6
Substitute n = 6 in (1), we get,
    (n -1) × d = 54
(6 - 1) × d = 54
5d = 54
d =
Thus n = 6 and d = .


(iii) Given an = 4, d = 2, Sn = -14, find n and a.
Here, an = 4, d = 2, Sn = -14.
an = a +(n – 1)d
4 = a +2(n – 1)
a = 4 – [2(n -1)] …………………… (1)
Sn =
-14 =
      =
Substitute a = 4 – [2(n -1)] from (1)
-14 = n[4 – 2(n – 1) + (n – 1)]
-14 = n[4 – (n – 1)]
 = 4n – n2 + n
n2 – 5n – 14 = 0
n2 – 7n + 2n – 14 = 0
n(n – 7) + 2(n - 7) = 0
(n - 7)(n+2) = 0
n = 7, n = -2
Number of terms cannot be negative thus n = 7 9; 9;
Substitute n = 7 in (1), we get,
a = 4 – [2 (7 – 1)]
= 4 – (2 × 6)
= 4 – 12 = -8
n = 7, a = -8
Thus n = 7 and a = -8.

(iv) Given a = 3, n = 8, S = 192, find d.
a = 3, n = 8, S = 192 find d
Sn =
192 =
192 = 4[(2 × 3) + (8 – 1)d]

192 = 4(6 + (8 – 1)d)
6 + 7d =
6 + 7d = 48
7d = 42
d = = 6
Thus d = 6.

(v) Given l = 28, S = 144, and there are total 9 items. Find a.
l = 28, S = 144, n = 9, a = ?
S = (a+l)
144 =
144 × = a + 28
32 = a + 28
a = 4
Hence a = 4.

Question-29

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

Solution:
The given A.P is 9, 17, 25, …
Sn = 636

In this A.P, a = 9

d = a2 – a1 = 17 – 9 = 8
Let n be the number of terms
Sn =
636 =
1272 =
        =
        = 8n2 + 10n
8n2 + 10n – 1272 = 0
Dividing throughout by 2, we get
4n2 + 5n – 636 = 0
4n2 + 53n – 48n - 636 = 0
n(4n + 53) – 12(4n + 53) = 0
(n - 12)(4n + 53) = 0
4n = -53 n = ; n = 12
Only one value n = 12 is admissible.
Thus the number of terms is 12.

Question-30

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

Solution:
a = 5, an = 45, Sn = 400
Sn = where l - the last term, a - first term and n - number of terms
400 =
       =
400 = 25n
n = = 16
Number of terms = 16
an = a + (n-1)d
45 = 5 +(16 – 1) d
45 – 5 = 15d
d =
Hence the number of terms and the common difference is 16 and .

Question-31

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

Solution:
In this problem,
a = 17, l = 350, d = 9
Let n be the no. of terms and Sn their sum
Let an = 350
an = a + (n – 1)d
350 = 17 + (n – 1) 9
350 – 17 = (n – 1) 9
= (n – 1)
(n – 1) = 37
n = 38
Sn =
    =
    = = 6973.
Thus the number of terms is 38 and its sum is 6973.

Question-32

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

Solution:
d = 7, a22 = 149
n = 22
As an = a + (n – 1)d
a22 = a + (n – 1) × 7
149 = a + 21 × 7
       = a + 147
a = 149 - 147
a = 2
Sn =
       =
       = 11 × 151
The sum of first 22 terms = 1661

Question-33

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

Solution:
a2 = 14
a3 = 18
a2 = 14 = a + (2-1)d
a + d = 14 …………………………(1)
a3 = 18
a + 2d = 18 ………………………….(2)
a + d = 14
a + 2d = 18
(1) – (2) -d = -4 d = 4
Substituting d = 4 in (1), we get,
a + d = 14
a + 4 = 14
a = 14 – 4 = 10
Sn =
S51 =
=
=
= = 5610.
The sum of first 51 terms = 5610.

Question-34

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

Solution:
S7 = 49
S17 = 289
Sn =
S7 =
49 =
49 × = (2a + 6d)
7= a + 3d ------------------- (1) (Dividing by 2)
S17 = [2a + (17 – 1) d]
289 = [2a + (17 – 1) d]
289 × = (2a + 16d)
      17 = a + 8d ……………………….. (2) (Dividing by 2)
(1) 7= a + 3d
(2) -(1)  5d = 10
d = 2
Substituting d = 2 in (1), we get,
a + 3d = 7
a + 3(2) = 7
a = 7 – 6
a = 1.
Sn =
     = =
     = n2
Hence the sum of first n terms is n2.

Question-35

Show that a1, a2, …, an, … form an A.P where an is defined as below:
(i) an = 3 + 4n (ii) a = 9 - 5n

Solution:
(i) Given, an = 3 + 4n   
When n = 1,
a1 = 3 + 4(1)= 3 + 4 = 7
When n = 2,
a2 = 3 + 4(2) = 3 + 8 = 11
When n = 3, a3 = 3 + 4(3) = 3 +12 = 15
Since a2 – a1 = 11 – 7 = 4 and a3 – a2 = 15 – 11 = 4.
i.e., ak+1 - ak is the same every time, the given list forms an A.P with the common difference (d) = 4
Sn =
S15 =
      =
      =
      =
      = = 525
This is an A.P with a = 7, d = 4, S15 = 525.

(ii) Given,  a = 9 - 5n
When n = 1,  a1 = 4
When n = 2, a2 = 9 - 10 = -1
When n = 3,  a3 = 9 – 15
            = -6
Since a2 – a1 = -1 – 4 = -5 and a3 – a2 = -6 + 1 = -5.
i.e., ak+1 - ak is the same every time, the given list forms an A.P with the common difference (d) = -5
Sn =
S15 =
      =
      =
      =
      = - 465.
This is an A.P with a = 4, d = -5, S15 = -465

Question-36

If the sum of the first n terms of an A.P is 4n – n2, 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:
Sn = (4n – n2)
a1 = S1 = 4(1) – (1)2
     = 4 – 1 = 3
S2 = 4(2) – 22 = 4,
S3 = 4(3) – 32
     = 12 – 9
     = 3
a2 = S2 – S1 = 4 – 3 = 1
a3 = S3 – S2 = 3 – 4 = -1
a1, a2, a3 is given by 3, 1, -1
Here a = 3, d = -2, n = 10
∴ a10 = a + (n – 1)d
         = 3 + 9 × (-2)
         = 3 – 18
         = -15
an = 3 + (n – 1) (– 2)
an = -2n + 5 = 5 – 2n
Thus a1, a2, a3 is given by 3, 1, -1, the 10th term is –15, and the nth term is 5 – 2n.

Question-37

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

Solution:
First term a = 6, d = 6, n = 40
Sum of the 40 positive integers divisible by 6
Sn =
S40 =
= 20[12 + (39 × 6)] = 20 [12 + 234]
= 20 × 246
∴ S40= 4920

Question-38

Find the sum of the first 15 multiples of 8.

Solution:
First term a = 8, d = 8, n = 15
Sum of 1st 15 multiples of 8:
Sn =
S15 =
     =
     =
     = 960

Question-39

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

Solution:
The list of odd numbers between 0 and 50: 1, 3, 5, …, 49.
First term (a) = 1
Common difference (d) = 2
Last term (l) = 49
Sn = ………………….. (1)
Last term is given by
an = 1 + (n – 1) × d
    = 1 + (n – 1) × 2
2n – 2 = 49 – 1
2n = 48 + 2
2n = 50
n = 25
Substituting n = 25 in (1), we get,
Sum = = 625.
The sum of odd numbers between 0 and 50 = 625

Question-40

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: `200 for the first day, `250 for the second day, `300 for the third day, etc., the penalty for each succeeding day being `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:
a1 = 200, a2 = 250, a3 = 300
(i.e.,) a = 200, d = 50
If n = 30,
Sn =
    =
    =
    =
    = 1850 × 15 = `27,750
If the contractor delays the work by 30 days, he has to pay `27,750 as penalty.

Question-41

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

Solution:
S7 = 700
d = `(-20) less than preceding
   = (-20)
a1 = a + d = a – 20
a2 = a - 40
a3 = a – 60
a7 = a – 140
Now
Sn =
S7 =
     =
700 =
a - 60= 100
a = 160
Values of the prizes (in `) are 160, 140, 120, 100, 80, 60, 40.

Question-42

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:
Hence the A.P will be of the form 1, 2, 3, 4, …, 12
1st term (a) = 1
Last term (l) = 12
Common difference (d) = 1
Sum of trees planted by one section of each class =
                                                                     =
                                                                     = 13 × 6
                                                                     = 78 .
Since there are 3 sections in each class the total number of trees planted = 78 × 3 = 234.
Total number of trees planted by the students = 234.

Question-43

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, … as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take p = 22/7)
 

 


Solution:
Length of semicircle with the centre at A = l1
Length of semicircle with the centre at B = l2
Length of semicircle with the centre at A = l3 and so on…
Applying the formula, length of semicircle =
As 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, … as shown in the figure
The total length of all the semicircles = π × (0.5) + π × (1.0) + π × (1.5) +…
= π [0.5 + 1 + 1.5 +…]
This is an AP with first term = 0.5 and d = 0.5
If n = 13,
Sn =
& S13 =
       = = = =
Total length of the thirteen consecutive semicircles = π ×
                                                                        = = 143 cm.

Question-44

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

 


Solution:
Number of logs in 1st row (a) = 20
Total number of logs = Sn = 200
d = (-1)
Sn =
200 =
      =
400 = n(41 – n)
(41 – n)n = 400
41n - n2 – 400 = 0
41n – n2 – 400 = 0
n2 – 41n + 400 = 0
n2 – 16n – 25n + 400 = 0
n(n – 16) – 25(n – 16) = 0
(n – 25)(n – 16) = 0
n = 16 (or) n = 25
Number of rows is either 16 or 25 (substituting for a and d)
If n = 16,
an = a + (n – 1) d
a16 = 20 + (16 – 1)(-1)
a16 = 20 - 15 = 5
Substituting for a and d
If n = 25, a25 = 20 +(25 – 1) (-1)
= 20 – 24 = -ve
Since the number of logs in the row cannot be negative (n 25)
Hence the number of rows = 16, and number of logs in the top row = 5
[(i.e.,) n = 16, a16 = 5]

Question-45

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:
To pick up the 1st potato the competitor has to run 2 × 5 m. Similarly for the 2nd potato
2 × 5 + 2 × 3 = 2 × (5 + 3)
Similarly for the 3rd potato the competitor has to run = 2 × (5 + 3 + 3)
If we write down the distance traveled by the competitor for each potato it can be given by
2 × 5, 2 × (5 + 3), 2 × (5 + 3 + 3) etc
(i.e.,) 2 times 5, 8, 11, …
a = 5, d = 3, n =10
Sn =
S10 =
       =
       = 5 (10 + 27)
       = 5 (37)
       = 185
Total distance traveled = 2 times (5 + 8 + 11 +…)
    = 2 × 185
    = 370 m.




Test Your Skills Now!
Take a Quiz now
Reviewer Name