In this post, you will find the NCERT solutions for class 10 maths ex 5.3. These solutions are based on the latest syllabus of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 5.3
1.
Find the sum of the following Aps:
(i)
2, 7, 12, …, to 10 terms.
(ii)
–37, –33, –29, …, to 12 terms.
(iii)
0.6, 1.7, 2.8, …, to 100 terms.
(iv) 1/15,
1/12, 1/10, …, to 11 terms.
Solution: (i) 2, 7, 12, …, to 10 terms.
Here, first term (a) = 2, common difference (d) = 7 – 2 = 5 and n = 10
Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n
terms of an AP, we have
Sn = (10/2)[4 + (10 – 1)5]
= 5(4 + 45)
= 5 × 49
= 245
(ii)
–37, –33, –29, …, to 12 terms.
Here, first term (a) = –37, common difference (d) = –33 – (–37) = 4 and n = 12
Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n
terms of an AP, we have
Sn = (12/2)[–74 + (12 – 1)4]
= 6(–74 + 44)
= 6 × (–30)
= –180
(iii)
0.6, 1.7, 2.8, …, to 100 terms.
Here, first term (a) = 0.6, common difference (d) = 1.7 – 0.6 = 1.1 and n = 100
Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n
terms of an AP, we have
Sn = (100/2)[1.2 + (100 –
1)1.1]
=
50(1.2 + 108.9)
= 50 ×
110.1
= 5505
(iv) 1/15,
1/12, 1/10, …, to 11 terms.
Here, first term (a) = 1/15, common difference (d) = 
Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n
terms of an AP, we have
2. Find the sums given below:
(i)
7 + 10½ + 14 + … + 84
(ii)
34 + 32 + 30 + … + 10
(iii)
–5 + (–8) + (–11) + … + (–230)
Solution:
(i) 7 + 10½ + 14 + … + 84
Here, first term (a) = 7, common difference (d) = 
Then, last term (l) =
an = 84
So, we have to find n first.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
[7 +
(n − 1) (3.5)] = 84
⇒ 7 +
3.5n − 3.5 = 84
⇒ 3.5n = 84 + 3.5 – 7
⇒ 3.5n = 80.5
⇒ n = 23
Therefore, there are 23
terms in the given AP.
Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of an AP, we have
S13 = (23/2)[7 + 84]
= (23/2)[91]
= 1046.5
(ii) 34 + 32 + 30 + … + 10
Here, first term (a) = 34, common difference (d) = 32 – 34 = –2
Let the number of terms in
the given AP be n.
Then, last term (l) =
an = 10
So, we have to find n first.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
[34
+ (n − 1) (−2)] = 10
⇒ 34
– 2n + 2 = 10
⇒ −2n = −26
⇒ n = 13
Therefore, there are 13
terms in the given AP.
Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of AP, we have
S13 = (13/2)[34 + 10]
= (13/2)[44]
= 286
(iii)
–5 + (–8) + (–11) + … + (–230)
Here, first term (a) = –5, common difference (d)
= –8 – (–5) = –8 + 5 = –3
Let the number of terms in
the given AP be n.
Then, last term (l) =
an = –230
So, we have to find n first.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
[−5
+ (n − 1) (−3)] = −230
⇒ −5
− 3n + 3 = −230
⇒ −3n = −228
⇒ n = 76
Therefore, there are 76
terms in the given AP.
Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of AP, we have
S76 = (76/2)[−5 − 230]
= 38 (−235)
= −8930
3.
In an AP:
(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.
(vi)
given a = 2, d = 8, Sn = 90, find n
and an.
(vii)
given a = 8, an
= 62, Sn
= 210, find n and d.
(viii)
given an = 4, d =
2, Sn = −14,
find n and a.
(ix)
given a = 3, n = 8, S = 192,
find d.
(x)
given l = 28, S = 144, and there are total 9
terms. Find a.
Solution:
(i) given a = 5, d = 3, an = 50, find n and Sn.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
an = 5 + (n − 1)(3)
⇒ 50
= 5 + 3n − 3
⇒ 48
= 3n
⇒ n = 16
Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n
terms of an AP, we have
S16 = (16/2)[10 + (16 – 1)3]
= 8(10 + 45)
= 8 × 55
= 440
Therefore, n = 16 and Sn = 440
(ii)
given a = 7, a13
= 35, find d and S13.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
a13= 7 + (13 − 1)d
⇒ 35
= 7 + 12d
⇒ 28
= 12d
⇒ d = 7/3
Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n
terms of an AP, we have
S13 = (13/2)[14 + (13 – 1)7/3]
= (13/2)(14 + 28)
= (13/2) × 42
= 273
Therefore, d = 7/3 and S13 = 273
(iii)
given a12 = 37, d =
3, find a and S12.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
a12= a + (12 − 1)3
⇒ 37
= a + 33
⇒ a = 4
Applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n
terms of an AP, we have
S12 = (12/2)[8 + (12 – 1)3]
= 6(8 + 33)
= 6 × 41
= 246
Therefore, a = 4 and S12 = 246
(iv)
given a3 = 15, S10 = 125, find d and a10.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
a3 = a + (3 − 1) (d)
⇒ 15
= a + 2d
⇒ a = 15 − 2d … (1)
Applying
formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n terms of an AP, we have
S10 = (10/2)[2a + (10 – 1)d]
⇒ 125 = 5(2a + 9d)
⇒ 25 = 2a + 9d … (2)
Putting the value of a from equation (1) in equation (2), we
have
25 =
2(15 − 2d) + 9d = 30 − 4d + 9d
⇒ 25
= 30 + 5d
⇒ –5
= 5d
⇒ d = −1
Putting
the value of d in equation (1), we
get
a =
15 – 2(–1)
a = 15
+ 2
a =
17
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
a10= a + (10 − 1)d
Putting
the values of a and d in the above equation, we get
a10 = 17 + 9 × (–1)
= 17
– 9
= 8
Therefore, d = −1 and a10 = 8
(v)
given d = 5, S9
= 75, find a and a9.
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S9 = (9/2)[2a + (9 – 1)5]
75 = (9/2)[2a + 40]
⇒ 150
= 18a + 360
⇒
−210 = 18a
⇒ a = −35/3
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
a9 = −35/3 + (9 − 1) (5)
= −35/3 + 40
= 85/3
Therefore, a = −35/3 and a9 = 85/3
(vi)
given a = 2, d = 8, Sn = 90, find n
and an.
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
90 = (n/2)[4
+ (n – 1)8]
⇒ 90 = (n/2)[4 + 8n – 8]
⇒ 90 = (n/2)[8n – 4]
⇒ 8n2 – 4n – 180 = 0
⇒ 2n2 – n – 45 = 0
⇒ 2n2 – 10n + 9n
– 45 = 0
⇒ 2n(n − 5) + 9(n − 5) = 0
⇒ (n − 5) (2n + 9) = 0
⇒ n = 5, −9/2
[We discard the negative
value of n because n cannot be negative or a fraction.]
Therefore, n = 5
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
a5 = 2 + (5 − 1) (8) = 2
+ 32 = 34
Therefore, n = 5 and an
= 34.
(vii)
given a = 8, an
= 62, Sn
= 210, find n and d.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
62 =
8 + (n − 1) (d)
⇒ 62
= 8 + nd − d
⇒ nd – d = 54
⇒ nd = 54 + d … (1)
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
210 = (n/2)[16
+ (n – 1)d]
210 = (n/2)[16
+ nd – d]
Putting the value of nd from equation (1) in the above
equation, we have
210 = (n/2)[16
+ 54 + d – d]
210 = (n/2)[70]
210 = n × 35
⇒ n = 6
Putting
the value of n in equation (1), we get
6d = 54 + d
⇒ 5d = 54
⇒ d = 54/5
Therefore, n = 6 and d = 54/5.
(viii)
given an = 4, d =
2, Sn = −14,
find n and a.
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we have
4
= a + (n − 1) (2)
⇒ 4
= a + 2n – 2
⇒ 6
= a + 2n
⇒ a = 6 − 2n … (1)
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
−14 = (n/2)[2a + (n
– 1)2]
−14 = (n/2)[2a + 2n
– 2]
−28 = n[2a + 2n
– 2]
Putting the value of a from equation (1) in the above
equation, we get
−28
= n[2(6 − 2n) + 2n – 2]
⇒ −28
= n(12 − 4n + 2n − 2)
⇒ −28
= n(10 − 2n)
⇒ 2n2 – 10n – 28 = 0
⇒ n2 – 5n – 14 = 0
⇒ n2 – 7n + 2n
– 14 = 0
⇒ n(n − 7) + 2(n − 7) = 0
⇒ (n + 2) (n − 7) = 0
⇒ n = −2, 7
[We
discard the negative value of n because it
cannot be negative.]
Therefore, n = 7.
Putting
the value of n in equation (1), we get
a = 6 − 2n = 6 – 2(7) = 6 – 14 = −8
Therefore, n = 7 and a = −8.
(ix)
given a = 3, n = 8, S = 192,
find d.
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
192
= (8/2)[6 + (8 − 1)d]
⇒ 192
= 4(6 + 7d)
⇒ 192
= 24 + 28d
⇒ 168
= 28d
⇒ d = 6
(x)
given l = 28, S = 144, and there are total 9
terms. Find a.
Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of an AP, we get
144
= (9/2)[a + 28]
⇒ 288
= 9[a + 28]
⇒ 32
= a + 28
⇒ a = 4
4. How many terms of the AP: 9, 17, 25, … must be taken to give a
sum of 636?
Solution:
Here, first term (a) = 9, common
difference (d) = 17 – 9 = 8 and Sn = 636
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
636
= (n/2)[18 + (n − 1) (8)]
⇒
1272 = n(18 + 8n − 8)
⇒
1272 = n(10 + 8n)
⇒ 1272 = 10n + 8n2
⇒ 8n2 + 10n – 1272 = 0
⇒ 4n2 + 5n – 636 = 0
Comparing equation 4n2 + 5n – 636
= 0 with the general form an2
+ bn + c = 0, we get
a = 4, b = 5 and c = −636
Applying quadratic
formula, 
and putting the values of a,
b and c, we get
⇒ 
⇒ 
⇒ 
[Here, we discard negative
value of n because n
cannot be negative.]
Therefore, n = 12
Hence, 12 terms of the given
sequence give a sum of 636.
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:
Here, first term (a) = 5, last
term (l) =
45 and Sn = 400
Applying formula, Sn = (n/2)[a + l] to find the sum of n terms of an AP, we get
400 = (n/2)[5 + 45]
400 = (n/2) × 50
400 = n × 25
⇒ n = 16
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
400 = (16/2)[10 + (16 – 1)d]
⇒ 400
= 8(10 + 15d)
⇒ 400
= 80 + 120d
⇒ 320
= 120d
⇒ d = 320/120
⇒ d = 8/3
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:
Here, first term (a) = 17, last term (l) =
350 and common difference (d) = 9
Using formula, an = a + (n – 1)d, to find the nth term of an AP, we get
350
= 17 + (n − 1)(9)
⇒ 350
= 17 + 9n − 9
⇒ 342
= 9n
⇒ n = 38
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S38 = (38/2)[34 + (38 – 1)9]
⇒ S38 =
19(34 + 333) = 6973
Therefore, there are 38
terms and their sum is equal to 6973.
7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
It is given that 22nd term is 149
It
means a22 = 149
Using formula, an = a + (n – 1)d, to find the nth term of an AP, we get
149
= a + (22 − 1)(7)
⇒ 149
= a + 147
⇒ a = 2
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
⇒ S22= (22/2)[4
+ (22 – 1)7)
⇒ S22=
11(4 + 147)
⇒ S22=
1661
Therefore, the sum of first
22 terms of the AP is 1661.
8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and
18 respectively.
Solution: It is given that the second and third terms of an AP are 14 and 18
respectively.
It means that a2 = 14 and a3 = 18
Using formula, an = a + (n – 1)d, to find the nth term of an AP, we get
14
= a + (2 − 1)d
⇒ 14
= a + d … (1)
And,
18 = a + (3 − 1)d
⇒ 18
= a + 2d … (2)
These are the linear
equations in two variables.
Using
equation (1), we get a = 14 − d
Putting
the value of a in equation (2), we get
18 =
14 – d + 2d
⇒ d = 4
Therefore,
common difference (d) = 4
Putting
the value of d in equation (2), we get
18
= a + 2(4)
⇒ a = 10
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S51 = (51/2)[20 + (51 – 1)4]
S51 = (51/2)[20 + 200]
S51 = (51/2)[220]
S51 = 51 × 110
S51 = 5610
Therefore, sum of first 51
terms of an AP is equal to 5610.
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: It is given that the sum of first 7 terms of an AP is 49 and the sum
of first 17 terms is 289.
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
49 = (7/2)[2a + (7 – 1)d]
⇒ 98
= 7(2a + 6d)
⇒ 14
= 2a + 6d
⇒ 7
= a + 3d
⇒ a = 7 − 3d … (1)
And, 289 = (17/2)[2a + (17 – 1)d]
⇒ 578
= 17(2a + 16d)
⇒ 34
= 2a + 16d
⇒ 17
= a + 8d … (2)
Putting the value of a from equation (1) in equation (2), we
get
17 =
7 − 3d + 8d
⇒ 10
= 5d
⇒ d = 2
Putting
the value of d in equation (1), we get
a = 7 − 3d
= 7
– 3(2)
= 7 – 6
= 1
Again applying formula, Sn = (n/2)[2a + (n – 1)d] to find the sum of n
terms of an AP, we have
Sn = (n/2)[2(1) + (n – 1)2]
Sn = (n/2)[2 + 2n – 2]
Sn = (n/2) × 2n
Sn = n2
Therefore,
the sum of n terms of the AP is n2.
10.
Show that a1, a2, …, an … form an AP where an is defined as below:
(i) an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first
15 terms in each case.
Solution:
(i) We
have to show that a1, a2
…, an … form
an AP where an = 3 + 4n
Let
us calculate the values of a1, a2 …, using an = 3 + 4n
a1 = 3 + 4(1) = 3 + 4 = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
So, the sequence is of the
form 7, 11, 15, 19, …
Let us find the difference
between two consecutive terms of the above sequence.
11 – 7 = 4, 15 – 11 = 4, 19
– 15 = 4
Since the difference between
two consecutive terms is the same which means that the terms a1, a2 …, an form an AP.
We have the sequence: 7, 11,
15, 19 …
Here, first term (a) = 7 and common difference (d) = 4
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S15 = (15/2)[14 + (15 – 1)4]
S15 = (15/2)[14 + 56]
S15 = (15/2) × 70
S15 = 15 × 35
S15 = 525
Therefore, the sum of the
first 15 terms of the AP is 525.
(ii) We have to show that a1, a2 …, an … form an AP where an = 9 – 5n
Let us calculate the values
of a1, a2 …, an … using an = 9 – 5n
a1 = 9 – 5(1) = 9 – 5 = 4
a2 = 9 – 5(2) = 9 – 10 = −1
a3 = 9 – 5(3) = 9 – 15 = −6
a4 = 9 – 5(4) = 9 – 20 = −11
So, the sequence is of the
form 4, −1, −6, −11, …
Let us find the difference
between two consecutive terms of the above sequence.
–1 – (4) = –5
–6 – (–1) = –6 + 1 = –5
–11 – (–6) = –11 + 6 = –5
Since the difference between
two consecutive terms is the same which means that the terms a1, a2 …, an form an AP.
We have the sequence: 4, −1,
−6, −11, …
Here, first term (a) = 4 and common difference (d) = −5
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S15 = (15/2)[8 + (15 – 1)(–5)]
S15 = (15/2)[8 – 70]
S15 = (15/2)[–62]
S15 = 15 × (–31)
S15 = –465
Therefore, the sum of the
first 15 terms of the AP is –465.
11.
If the sum of the first n terms of an
AP 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: It is given that the sum of the first n terms of an AP is (4n – n2).
It means Sn = 4n – n2
Let us calculate S1 and S2 using Sn
= 4n – n2
S1 = 4(1) – (1)2 = 4 – 1 = 3
S2 = 4(2) – (2)2 = 8 – 4 = 4
First term (a) = S1=
3 … (1)
Now, let us find the common
difference.
We can write any AP in the
form of general terms like a, a + d,
a + 2d …
We have calculated that S2 = 4
Therefore, we can say that a + (a
+ d) = 4
⇒ 2a + d = 4
Putting the value of a from equation (1), we get
⇒ 2(3) + d = 4
⇒ 6 + d = 4
⇒ d = –2
Using
formula, an = a + (n
– 1)d, to find the nth term of an
AP, we get
a2 = a + (2 – 1)d
= 3 + (2 – 1)(–2)
= 3 – 2 = 1
a3 = a + (3 – 1)d
= 3 + (3 – 1)(–2)
= 3 – 4 = –1
a10 = a + (10 – 1)d
= 3 + (10 – 1) (–2)
= 3 – 18 = –15
an = a + (n – 1)d
= 3 + (n
– 1)(–2)
= 3 – 2n + 2
= 5 – 2n
12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The first 40 positive integers divisible by 6
are 6, 12, 18, 24 … up to 40 terms.
Therefore, we have to find
the sum of 40 terms of the above sequence.
Here, first term (a) = 6, common difference (d) = 12 – 6 = 6 and n = 40
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S40 = (40/2)[12 + (40 – 1)6]
= 20(12 + 39 × 6)
= 20(12 + 234)
= 20 × 246 = 4920
13.
Find the sum of the first 15 multiples of 8.
Solution: The first 15 multiples of 8 are 8, 16, 24, 32 … up to 15 terms.
Therefore, we have to find
the sum of 15 terms of the above sequence.
Here, first term (a) = 8, common difference (d) = 16 – 8 = 8 and n = 15.
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S15 = (15/2)[16 + (15 – 1)8]
S15 = (15/2)[16 + 14 × 8]
S15 = (15/2)[16 + 112]
S15 = (15/2)128
S15 = 15 × 64
S15 = 960
14.
Find the sum of the odd numbers between 0 and 50.
Solution: The odd numbers between 0 and 50 are 1, 3, 5, 7, …, 49
It is an AP because the difference between two
consecutive terms is the same.
Here,
first term (a) = 1, common difference
(d) = 3 – 1 = 2 and last term (l) = 49
We do not know that how many
odd numbers are there between 0 and 50.
Therefore, we have to find n first.
Using formula, an = a + (n
− 1) d, to find the nth term of an AP, we get
49 = 1 + (n − 1)2
⇒ 49 − 1 = 2n – 2
⇒ 48 = 2n − 2
⇒ 2n = 50
⇒ n = 25
Applying formula, Sn = (n/2)[a
+ l] to find the sum of n terms of an AP, we get
S25 = (25/2)[1 + 49]
S25 = (25/2)[50]
S25 = 25 × 25
S25 = 625
Therefore, the sum of the odd numbers between
0 and 50 is 625.
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: Penalty for the first day = Rs 200, penalty for the second day = Rs
250 and
penalty for the third day =
Rs 300
It is given that the penalty for each
succeeding day is Rs 50 more than the preceding day.
It forms an arithmetic progression because the
difference between any two consecutive terms is the same.
We have to find how much money the contractor
has to pay as penalty, if he has delayed the work by 30 days.
So, we have an AP of the form 200, 250, 300,
350 … up to 30 terms.
Here, first term (a) = 200, common difference (d)
= 50 and n = 30
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
Sn = (30/2)(400 + (30 – 1)50)
⇒ Sn =
15(400 + 29 × 50)
⇒ Sn = 15(400
+ 1450) = 27750
Therefore, the penalty for
30 days is Rs 27,750.
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: It is given that the sum of seven cash prizes is equal to Rs 700.
And each prize is Rs 20 less
than its preceding prize.
Let the value of first prize
be Rs a.
Then the value of second
prize = Rs (a − 20)
The value of third prize =
Rs (a − 40)
So, we have the sequence of
the form:
a, a − 20, a − 40, a – 60 …
It is an arithmetic progression because the
difference between two consecutive terms is the same.
Here, first term = a, common difference (d) = (a – 20) – a = –20
n = 7 (Because there are
total of seven prizes)
It is given that S7 = Rs 700
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S7 = (7/2)[2a + (7 – 1)(–20)]
⇒ 700 = (7/2)[2a – 120]
⇒ 200 = 2a – 120
⇒ 320 = 2a
⇒ a = 160
Therefore, the value of
first prize = Rs 160
The value of second prize =
160 – 20 = Rs 140
The value of third prize =
140 – 20 = Rs 120
The value of fourth prize =
120 – 20 = Rs 100
The value of fifth prize =
100 – 20 = Rs 80
The value of sixth prize =
80 – 20 = Rs 60
The value of seventh prize =
60 – 20 = Rs 40
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 two trees and so on till Class XII. There are three sections of
each class. How many trees will be planted by the students?
Solution: It is given that the number of trees planted by each class is equal
to class number. There are three sections of each class.
The number of trees planted by class I =
number of sections × 1 = 3 × 1 = 3
The number of trees planted by class II =
number of sections × 2 = 3 × 2 = 6
The number of trees planted by class III =
number of sections × 3 = 3 × 3 = 9
Therefore, we have sequence of the form 3, 6,
9 … up to 12 terms.
To find the total number of trees planted by
all the students, we have to find the sum of the sequence 3, 6, 9, 12 … up to 12
terms.
Here, first term (a) = 3, common difference (d) = 6 – 3 = 3 and n = 12
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S12 = (12/2)[6 + (12 – 1)3]
S12 = 6(6 + 33)
S12 = 6 × 39
S12 = 234
Hence, all the students planted 234
trees.
18.
A spiral is made up of successive semicircles, with centres alternatively 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:
Length of semicircle = 
Length of semicircle of
radii 0.5 cm = π(0.5) cm
Length of semicircle of
radii 1.0 cm = π(1.0) cm
Length of semicircle of
radii 1.5 cm = π(1.5) cm
Therefore, we have the sequence
of the form:
Ï€(0.5), Ï€(1.0), Ï€(1.5) … up to 13 terms [Since there are total thirteen semicircles.]
To find the total length of the spiral, we have
to find the sum of the sequence Ï€(0.5), Ï€(1.0), Ï€ (1.5) … up to 13 terms.
Total length of the spiral = Ï€(0.5) + Ï€(1.0) + Ï€(1.5) … up to 13 terms
Total length of the spiral =
Ï€(0.5
+ 1.0 + 1.5) … up to 13 terms … (1)
The sequence 0.5, 1.0, 1.5 …
up to 13 terms is an arithmetic progression.
Let us find the sum of this
sequence.
Here, first term (a) = 0.5, common difference (d)
= 1.0 – 0.5 = 0.5 and n = 13
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S13 = (13/2)[1 + (13 – 1)0.5]
S13 = 6.5[1 + 6]
S13 = 6.5 × 7
S13 = 45.5
Therefore, 0.5 + 1.0 + 1.5 +
2.0 … up to 13 terms = 45.5
Putting this value in
equation (1), we get
Total length of the spiral = Ï€(0.5 + 1.5 + 2.0 + … up to 13 terms)
= π(45.5) = (22/7)(45.5) = 143 cm
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?
The number of logs in the
next row = 19
The number of logs in the next to next row =
18
Therefore, we have the
sequence of the form 20, 19, 18 …
Here, first term (a) = 20, common difference (d) = 19 – 20 = –1
We have to find that how
many rows make total of 200 logs. Let there are a total of n rows.
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
200 = (n/2)[40 + (n – 1) (–1)]
⇒ 400 = n(40 – n + 1)
⇒ 400 = 40n – n2 + n
⇒ n2 – 41n + 400 = 0
It is a quadratic equation.
We can factorize it to solve
the equation.
⇒ n2 − 25n − 16n + 400 = 0
⇒ n(n − 25) – 16(n − 25) = 0
⇒ (n − 25) (n − 16)
⇒ n = 25, 16
We discard n
= 25 because we cannot have more than 20 rows in the sequence.
The sequence is of the form: 20, 19, 18 …
Here, n = 16 which means 16 rows make a total number of logs equal to
200.
We also have to find the number of logs in the
16th row.
Applying formula, Sn = (n/2)[a + l] to
find the sum of n terms of an AP, we
get
200
= 8(20 + l)
⇒ 200
= 160 + 8I
⇒ 40
= 8l
⇒ l = 5
Therefore, there are 5 logs
in the topmost row and there are a total of 16 rows.
20.
In a potato race, a bucket is placed at the starting point, which is 5 metres
from the first potato, and the other potatoes are placed 3 metres 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: The distance of first potato from the starting point = 5 metres
Therefore, the distance covered by competitor
to pick up first potato and put it in bucket = 5 × 2 = 10 metres
The distance of second potato from the
starting point = 5 + 3 = 8 metres
[All the potatoes are 3 metres apart from each
other.]
Therefore, the distance covered by competitor
to pick up 2nd potato and put it in bucket = 8 × 2= 16 metres
The distance of third potato from the starting
point = 8 + 3 = 11 metres
Therefore, the distance covered by competitor
to pick up 3rd potato and put it in bucket = 11 × 2 = 22 metres
Therefore, we have a sequence of the form: 10,
16, 22 … up to 10 terms.
[There are a total of ten terms because there
are ten potatoes.]
To calculate the total distance covered by the
competitor, we have to find:
10 + 16 + 22 + … up to 10
terms.
Here, first term (a) = 10, common difference (d)
= 16 – 10 = 6 and n = 10
Applying formula, Sn = (n/2)[2a + (n
– 1)d] to find the sum of n terms of an AP, we have
S10 = (10/2)[20 + (10 – 1)6]
S10 = 5(20 + 54)
S10 = 5 × 74
S10 = 370
Therefore, the total distance covered by the
competitor is 370 metres.