Bihar Board Class 10th Maths Chapter 5 Arithmetic Progressions Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1

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 Rs 15 for the first km and Rs 8 for each additional km.
(ii) The amount of air present in the cylinder when a vacuum 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 Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) According to the statement, the fare for journey of 1 km, 2 km, 3 km, 4 km, … are respectively Rs 15, Rs (15 + 8), Rs (15 + 2 × 8), Rs (15 + 3 × 8), …
i.e., 15, 23, 31, 39, …
Since here the terms continually increase by the same number 8, so the list forms an AP.

(ii) Let the amount of air present in the cylinder be x units.
∴ According to the statement, the list giving the air present in the cylinder is given by

∴ These numbers do not form an AP.

(iii) According to the question, the cost of digging a well for the first metre and its succeeding metres in rupees is given by 150, 200, 250, 300, …
Since here the terms continually increase by the same number 50, so the list forms an AP.

(iv) According to the question, the amount of money in the account in the first year and its succeeding years in rupees is given by
10000, 10000(1 + 8100), 10000(1 + 8100)2, …………….
i.e; 10000, 10000 × 108100, 10000 × 108100 × 108100, …………..
i.e; 10000, 10800, 11664, ……………
Since 10800 – 10000 ≠ 11664 – 10800, therefore
the list does not form an AP.

Question 2.
Write first four terms of the AP, when the first term a and 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 = 12
(v) a = – 1.25, d = – 0.25
Solution:
We know that if the first term is a and the common difference is d, then
a, a + d, a + 2d, a + 3d, … represents an AP for different values of a and d.
(i) Putting a = 10 and d = 10 in a, a + d, a + 2d, a + 3d, …, we get the required AP as 10, 10 + 10, 10 + 20, 10 + 30, …….
i.e., 10, 20, 30, 40, …….

(ii) Putting a = – 2 and d = 0 in a, a + d, a + 2d, a + 3d, …, we get the required AP as – 2, – 2 + 0, – 2 + 2 × 0,- 2 + 3 × 0, …
i.e.,- 2, – 2, – 2, – 2, ……….

(iii) Putting a – 4 and d = – 3 in a, a + d, a + 2d, a + 3d, ………….., we get the required AP as 4, 4 – 3, 4 – 6, 4 – 9, …
i.e., 4, 1, – 2, – 5, ……..

(iv) Putting a = – 1 and d = 12 in a, a + d, a + 2d, a + 3d, …, we get the required AP as – 1,- 1 + 12, – 1 + 22, – 1 + 32, …………..
i.e; – 1, −12, 0, 12, ……………

(v) Putting a = – 1.25 and d = – 0.25 in a, a + d, a + 2d, a + 3d, …, we get the required AP as – 1.25, – 1.25 – 0.25, – 1.25 – 0.50, – 1.25 – 0.75, …
i.e; – 1.25, – 1.50, – 1.75, – 2.00, …….

Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, – 1, – 3, …
(ii) – 5, – 1, 3, 7, …
(iii) 13, 53, 93, 133, …………..
(iv) 0.6, 1.7, 2.8, 3.9 …
Solution:
(i) The given AP is 3, 1, – 1, – 3, …
Clearly, a = 3 and d = 1 – 3 = – 2.

(ii) The given AP is – 5, – 1, 3, 7, …..
Clearly, a = – 5 and d = – 1 – (- 5) = – 1 + 5 = 4

(iii) The given AP is 13, 53, 93, 133, ………..
Clearly, a = 13 and d = 53 – 13 = 43

(iv) The given AP is 0.6, 1.7, 2.8, 3.9, ………
Clearly, a = 0.6 and d, = 1.7 – 0.6 = 1.1

Question 4.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, …….
(ii) 2, 52, 3, 72, ……………..
(iii) – 1.2, – 3.2, – 5.2, – 7.2, ……….
(iv) – 10, – 6, – 2, 2, ……..
(v) 3, 3 + 2–√, 3 + 22–√, 3 + 32–√, ………….
(vi) 0.2, 0.22, 0.222, 0.2222, ……….
(vii) 0, – 4, – 8, – 12, …………..
(ix) 1, 3, 9, 27, ……….
(x) a, a2, a3, a4, ……..
(xi) 2–√, 8–√, 18−−√, 32−−√, …………..
(xii) 3–√, 6–√, 9–√, 12−−√, …………..
(xiii) 3–√, 6–√, 9–√, 12−−√, …………..
(xiv) 12, 32, 52, 72, …………..
(xv) 12, 52, 72, 73, ………
Solution:
(i) Here, a2 – a1 = 4 – 2 = 2
and a3 – a2 = 8 – 4 = 4
⇒ a2 – a1 ≠ a3 – a2
Thus, 2, 4, 8, 16, ……….. do not form an AP.

(ii) Here, a2 – a1 = 52 – 2 = 5−42 = 12,
a3 – a2 = 3 – 52 = 6−52 = 12,
and a4 – a3 = 72 – 3 = 7−62 = 12
i.e., an+1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = 12
The next three terms after the last given term are:
72 + 12 = 7+12 = 82 = 4,
4 + 12 = 412 = 92,
and 92 + 12 = 9+12 = 102 = 5.

(iii) Here, a2 – a1 = – 3.2 – (- 1.2) = – 3.2 + 1.2 = – 2,
a3 – a2 = – 5.2 – (- 3.2) = – 5.2 + 3.2 = – 2,
and a4 – a3 = – 7.2 – (- 5.2) = – 7.2 + 5.2 = – 2
i.e., an+1 + 1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = – 2
The next three terms after the last given term are:
– 7.2 – 2 = – 9.2,
– 9.2 – 2 = – 11.2,
and – 11.2 – 2 = – 13.2.

(iv) Here, a2 – a1 = – 6 – (- 10) = – 6 + 10 = 4,
a3 – a2 = – 2 – (- 6) = – 2 + 6 = 4,
and a4 – a3 = 2 – (- 2) = 2 + 2 = 4
i.e., an+1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = 4
The next three terms after the last given term are:
2 + 4 = 6,
6 + 4 = 10, and 10 + 4 = 14.

(v) Here, a2 – a1 = (3 + 2–√) – 3 = 2–√
a3 – a2 = (3 + 22–√) – (3 + 2–√) = 2–√
and a4 – a3 = (3 + 32–√) – (3 + 22–√) = 2–√
i.e., a4 – a3 = (3 + 32–√) – (3 + 22–√) = 2–√
i.e; an+1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = 2–√
The next three terms after the last given term are:
(3 + 32–√) + 2–√ = 3 + 42–√,
(3 + 42–√) + 2–√ = 3 + 52–√,
and (3 + 52–√) + 2–√ = 3 + 62–√.

(vi) Here, a2 – a1 = 0.22 – 0.2 – 0.22 – 0.20 = 0.02
and a3 – a2 = 0.222 – 0.22
= 0.222 – 0.220 = 0.002
∴ a2 – a1 ≠ a3 – a2
Thus, the given list of numbers does not form an AP.

(vii) Here, a2 – a1 = – 4 – 0 = – 4,
a3 – a2 = – 8 – (- 4) = – 8 + 4 = – 4,
and a4 – a3 = – 12 – (- 8) = – 12 + 8 = – 4.
i.e., an+1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = – 4
The next three terms after the last given term are:
– 12 + (- 4) = – 12 – 4 = – 16,
– 16 + (- 4) = – 16 – 4 = – 20,
and – 20 + (- 4) = – 20 – 4 = – 24.

(viii)

i.e., an+1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = 0
The next three terms after the last given term are
– 12, – 12 and – 12.

(ix) Here, a2 – a1 = 3 – 1 = 2
and a3 – a2 = 9 – 3 = 6
So, a2 – a1 ≠ a3 – a2
Thus, the given list of numbers does not form an A.P.

(x) Here, a2 – a1 = 2a – a = a,
a3 – a2 = 3a – 2a = a,
and a4 – a3 = 4a – 3a = a
i.e., an+1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = a
The next three terms after the last given term are:
4a + a = 5a,
5a + a = 6a
and 6a + a = 7a.

(xi) a2 – a1 = a2 – a = a(a – 1)
and a3 – a2 = a3 – a2 = a2(a – 1)
So, a2 – a1 ≠ a3 – a2
Thus, the given list of numbers does not form an AP.

(xii) Here, a2 – a1 = 8–√ – 2–√ = 4×2−−−−√ – 2–√
= 22–√ – 2–√ = 2–√,
a3 – a2 = 18−−√ – 8–√ = 9×2−−−−√ – 4×2−−−−√
= 32–√ – 22–√ = 2–√
and a4 – a3 = 32−−√ – 18−−√
= 16×2−−−−−√ – 9×2−−−−√
= 42–√ – 32–√ = 2–√
i.e; an+1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = 2–√
The next three terms after the last given term are

(xiii)

Thus, the given list of numbers does not form an AP.

(xiv) Here, a2 – a1 = 32 – 12 = 9 – 1 = 8,
and a3 – a2 = 52 – 32 = 25 – 9 = 16
So, a2 – a1 ≠ a3 – a2
Thus, the given list of numbers does not form an AP.

(xv) Here, a2 – a1 = 52 – 12 = 25 – 1 = 24,
a3 – a2 = 72 – 52 = 49 – 25 = 24,
and a4 – a3 = 73 – 72 = 73 – 49 = 24
i.e; an+1 – an is same every time. So the given list of numbers forms an AP.
Here, common difference d = 24
The next three terms after the last given term are:
73 + 24 = 97,
97 + 24 = 121,
and 121 + 24 = 145.

BSEB Bihar Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2

Question 1.
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 AP:

Solution:
Blanks may be filled as under:
(i) an – a + (n – 1)d = 7 + (8 – 1)3
= 7 + 7 × 3 = 7 + 21 = 28

(ii) an = a + (n – 1)d So, 0 = – 18 + (10 – 1)d
or 18 = 9d or d = 189 = 2

(iii) an = a + (n – 1)d So, – 5 = a + (18 – 1)(- 3)
or – 5 = a + 17(- 3) or – 5 = a – 51
or a = – 5 + 51 = 46

(iv) an = a + (n – 1)d So, 3.6 = – 18.9 + (n – 1)25
or 3.6 + 18.9 = (n – 1)2.5 or 22.5 = (n – 1)2.5
or n – 1 = 22.52.5 or n – 1 = 9
or n = 9 + 1 = 10

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

Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the 10, 7, 4, ………………. is
(A) 97
(B) 77
(C) – 77
(D) – 87

(ii) 11th term of the – 3, – 12 + 3 = −1+62 = 52 and n = 11.
We know that an = a + (n – 1)d
∴ a11 = – 3 + (11 – 1)52
= – 3 + 10 × 52 = – 3 + 25 = 22
So, (B) is the correct choice.

Question 3.
In the following APs, find the missing terms in the boxes:
(i) 2, , 26

(ii)
, 13, , 3
(iii) 5, , , 912
(iv) – 4, , , , , 6
(v) , 38, , , , – 22.
Solution:
(i) Let 2,  and 26 be a, (a + d) and (a + 2d) respectively.
∴ a = 2 and a + 2d = 26
So, 2 + 2d = 26 or 2d = 26 – 2 = 24
or d = 242 = 12
Thus, the missing term = a + d = 2 + 12 = 14.

(ii) Let , 13,  and 3 be a, a + d, a + 2d and a + 3d respectively.
∴ a + d = 13 ……………. (1)
and a + 3d = 3 ……………. (2)
So, (2) – (1) gives 2d = – 10 or d = – 5
Putting d = – 5 in (1), we get
a – 5 = 13 or a = 13 + 5 = 18
∴ The missing terms are a, i.e., 18
and a + 2d, i.e., 18 + 2(- 5) = 18 – 10 = 8.

(iii) Let 5, ,  and 912 be a, a + d, a + 2d and a + 3d respectively.
∴ a = 5
 and a + 2d, i.e; 5 + 2 × 32 = 5 + 3 = 8.

(iv) Let – 4, ,, ,  and 6 be a, a + d, a + 2d, a + 3d, a + 4d and a + 5d respectively.
∴ a = – 4 ………………. (1)
and a + 5d = 6 ……………… (2)
So, (2) – (1) gives 5d = 10 or d = 2
The missing terms are
a + d = – 4 + 2 = – 2,
a + 2d = – 4 + 2 × 2 = – 4 + 4 = 0,
a + 3d = – 4 + 3 × 2 = – 4 + 6 = 2,
and a + 4d = – 4 + 4 × 2 = – 4 + 8 = 4.

(v) Let , 38, , ,  and – 22 be a, a + d, a + 2d, a + 3d, a + 4d and a + 5d respectively.
∴ a + d = 38 ……….. (1)
and a + 5d = – 22 ……………. (2)
So, (2) – (1) gives 4d = – 60 or d = – 15
Putting d = – 15 in (1), we get
a – 15 = 38 or a = 38 + 15 = 53
∴ The missing terms are a, i.e., 53,
a + 2d, i.e., 53 + 2 × (- 15) = 53 – 30 = 23,
a + 3d, i.e., 53 + 3 × (- 15) = 53 – 45 = 8,
and a + 4d, i.e., 53 + 4 × (- 15) = 53 – 60 = – 7

Question 4.
Which term of the AP: 3, 8, 13, 18, …, is 78?
Solution:
We have:
a = 3, d = 8 – 3 – 5.
Let 78 be the nth term of the given AP. Then,
an = 78 or a + (n – 1 )d = 78
∴ 3 + (n – 1)5 = 78 or (n – 1)5 = 78 – 3
or 5(n – 1) = 75 or n – 1 = 15
or n = 15 + 1 i.e., n = 16
Thus, 78 is the 16th term of the given AP.

Question 5.
Find the number of terms in each of the following APs:
(i) 7, 13, 19, ………………., 205
(ii) 18, 1512, 13, ………………, – 47
Solution:
(i) Clearly, it forms an AP with first term a – 3 and common difference d = 13 – 7 = 6.
Let there be n terms in the given AP.
Then, nth term = 205.
So, a + (n – 1)d = 205 or 7 + (n – 1)6 = 205
or 6(n – 1) = 205 – 7 or 6(n – 1) = 198
or n – 1 = 1986 = 33 or n = 33 + 1 = 34
Thus, the given AP contains 34 terms.

(ii) Let there be n terms in the given AP.
18, 15 12, 13, …, – 47. Clearly, it forms an AP with first term a = 18 and common difference d = 1512 – 18 = 312 – 18 = 31−362 = −52. Then, nth term = – 47.
So, a + (n – 1)d = – 47 or 18 + (n – 1)(- 52) = – 47
or (- 52) (n – 1) = – 47 – 18 = – 65
or n – 1 = – 65 × −25 = – 13 × (- 2) = 26
Thus, there are 27 terms in the given AP.

Question 6.
Check whether – 150 is a term of the AP: 11, 8, 5, 2, …
Solution:
Here, a2 – a1 = 8 – 11 = – 3
a3 – a2 = 5 – 8 = – 3
a4 – a3 = 2 – 5 = – 3
As an+1 – an is same every time, so the given list of numbers is an AP.
Now, a = 11, d = – 3.
Let – 150 be the nth term of the given AP.
we know that an = a + (n – 1)d
So, – 150 = 11 + (n – 1)(- 3)
or – 3(n – 1) = – 150 – 11 = – 161
or n – 1 = 1613 or n = 1613 + 1 = 1643
But n should be a positive integer. So, our assumption was wrong and so – 150 is not a term of the given AP.

Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
Let a be the first term and d be the common difference of AP.
Now, an = a + (n – 1)d
∴ a11 = a + 10d = 38 …………….. (1)
and a16 = a + 15d = 73 …………….. (2)
Subtracting (1) from (2), we get
5d = 35 or d = 355 = 7
and then from (1),
a + 10 × 7 = 38
or a = 38 – 70 = – 32
∴ a31 = a + 30d = – 32 + 30 × 7
= – 32 + 210 = 178

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Let a be the first term and d be the common difference of AP.
Now, an = a + (n – 1)d
∴ a3 = a + 2d = 12 …………… (1)
and a50 = a + 49d = 106 …………….. (2)
Subtracting (1) from (2), we get
47d = 94 or d = 9447 = 2
and then from (1),
a + 2 × 2 = 12 or a = 12 – 4 = 8
∴ a29 = a + 28d = 8 + 28 × 2
= 8 + 56 = 64

Question 9.
If the 3rd and 9th terms of an AP are 4 and – 8 respectively, which term of this AP is 0?
Solution:
Let a be the first term and d be the common difference of AP.
a3 = a + 2d = 4 ………….. (1)
a9 = a + 8d = – 8 ………….. (2)
Subtracting (1) from (2), we get
6d = – 12 or d = −126 = – 2
and then from (1),
a + 2 × (- 2) = 4 or a + 4 + 4 = 8
Let an = 0, So, a + (n – 1)d = 0
or 8 + (n – 1)(- 2) = 0 or (n – 1)(- 2) = – 8
or n – 1 = −8−2 = 4 or n = 4 + 1 = 5
Thus, 5th term of the AP is 0.

Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Let a be the first term and d be the common difference of AP.
It is given that a17 – a10 = 7 or (a + 16d) – (a + 9d) = 7
or 7d = 7 or d = 1
Thus, the common difference is 1.

Question 11.
Which term of the AP : 3, 15, 27, 39, … will be 132 more than its 54th term?
Solution:
Here, a = 3, d = 15 – 3 = 12. Then,
a54 = a + 53d = 3 + 53 × 12
= 3 + 636 = 639
Let an be 132 more than its 54th term
i.e., an = a54 + 132 = 639 + 132 = 771
So, a + (n – 1)d = 771 or 3 + (n – 1)12 = 771
or 12(n – 1) = 771 – 3 or 12(n – 1) = 768
or n – 1 = 76812 = 64 or n = 64 + 1 = 65
Thus, 65th term of the AP is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let the two APs be a1, a2, a3, …………… an, …………. and b1, b2, b3, …………, bn, ………….
Also, let d be the common difference of each of the two APs. Then,
an = a1 + (n – 1 )d and bn = b1 + (n – 1)d
or an – bn = [a1 + (n – 1)d] – [b1 + (n – 1)d]
or an – bn = a1 – b1 for all natural numbers n
or a100 – b100 = a1 – b1 = 100 [Given]
Now, a1000 – b1000 = a1 – b1
So, a1000 – b1000 = 100 [∵ a1 – b1 = 100]
Hence, the difference between 1000th terms is the same as the difference between 100th terms, i.e., 100.

Question 13.
How many three-digit numbers are divisible by 7?
Solution:
105 is the first three-digit number divisible by 7 and 994 is the last 3-digit number divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119, …, 994.
Clearly, it forms an AP with first term a = 105
and common difference d = 112 – 105 = 7.
Let there be n terms in the AP. Then, nth term = 994.
As an = a + (n – 1)d, therefore
105 + (n – 1)7 = 994 or 7(n – 1) = 994 – 105
or 7(n – 1) = 889 or n – 1 = 8897 = 127
or n = 127 + 1 = 128
Hence, there are 128 numbers of three digit which are divisible by 7.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
We observe that 12 is the first natural number between 10 and 250, which is a multiple of 4 (i.e., divisible by 4).
Also, when we divide 250 by 4, the remainder is 2. Therefore, 250 – 2 = 248 is the largest natural number divisible by 4 (i.e., multiple of 4) and lying between 10 and 250.
Thus, we haye to find the number of terms in an AP with first term = 12, last term = 248, and common difference = 4 (as the numbers are divisible by 4).
Let there be n terms in the AP. Then
an = 248 So, 12 + (n – 1)4 = 248
or 4(n – 1) = 248 – 12 or 4(n – 1) = 236
or n – 1 = 2364 = 59 or n = 59 + 1 = 60
Hence, there are 60 multiples of 4 between 10 and 250.

Question 15.
For what value of n, are the nth terms of the APs: 63, 65, 67, … and 3, 10, 17, … are equal?
Solution:
If nth terms of the APs 63, 65, 67, … and 3, 10, 17, … are equal, then
63 + (n – 1)2 = 3 + (n – 1)7
[∵ In 1st AP, a = 63, d = 65 – 63 = 2 and in 2nd AP, a = 3, d = 10 – 3 = 7]
or 7(n – 1) – 2(n – 1) = 63 – 3 or (n – 1)(7 – 2) = 60
or 5(n – 1) = 60 or n – 1 = 605 = 12
or n = 12 + 1 = 13
Hence, the 13th terms of the two given APs are equal.

Question 16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
Let a be the first term and d be the common difference of AP.
Here, a3 = 16 and a7 – a5 = 12
So, a + 2d = 16 ……….. (1)
and (a + 6d) – (a + 4d) = 12 or 2d = 12
or d = 6 ……………. (2)
Using (2) in (1), we get
a + 2 × 6 = 16. or a = 16 – 12 = 4
Thus, the required AP is 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, … i.e., 4, 10, 16, 22, ……

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ………….. 253.
Solution:
We have:
l = Last term = 253
and d = Common difference = 8 – 3 = 5
∴ 20th term from the end = l – (20 – 1)d = l – 19d
= 253 – 19 × 5
= 253 – 95 = 158

Question 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.
Solution:
Let a be the first term and d be the common difference of AP.
Here, a4 + a8 = 24 or (a + 3d) + (a + 7d) = 24
or 2a + 10d = 24 or a + 5d = 12 ……….. (1)
and a6 + a10 = 44 or (a + 5d) + (a + 9d) = 44
or 2a + 14d = 44 or a + 7d = 22 ……… (2)
Subtracting (1) from (2), we get
2d = 10 or d = 5
and then from (1),
a + 25 = 12 or a = – 13
The first three terms are a, (a + d) and (a + 2d)
Putting values of a and d, we get – 13, (- 13 + 5) and (- 13 + 2 × 5)
i.e., – 13, – 8 and – 3.

Question 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?
Solution:
The annual salaries drawn by Subba Rao in the years 1995, 1996, 1997, etc. are Rs 5000, Rs 5200, Rs 5400, …, Rs 7000.
The list of these numbers is 5000, 5200, 5400, …, 7000.
It forms an AP. [∵ a2 – a1 = a3 – a2 = 200]
Let an = 7000
So, 7000 = a + (n – 1)d
or 7000 = 5000 + (n – 1)(200)
or 200(n – 1) = 7000 – 5000
or n – 1 = 2000200 = 10 or n = 10 + 1 = 11
Thus, in the 11th year (i.e., in 2005) of his service, Subba Rao drew an annual salary of Rs 7000.

Question 20.
Ramkali saves 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.
Solution:
Ramkali’s savings in the subsequent weeks are respectively Rs 5, Rs 5 + Rs 1.75, Rs 5 + 2 × Rs 1.75, Rs 5 + 3 × Rs 1.75, …
And in the nth week her savings will be Rs 5 + (n – 1) × Rs 1.75
or 5 + (n – 1) × 1.75 = 20.75
or (n – 1) × 1.75 = 20.75 – 5 = 15.75
∴ n – 1 = 15.751.75 = 9
or n = 9 + 1 = 10

BSEB Bihar Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 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.
Solution:
(i) Let a be the first term and d be the common difference of the given AP. Then, we have:
a = 2 and d = 7 – 2 = 5
We have to find the sum of 10 terms of the given AP.
Putting a = 2, d = 5 and n = 10 in Sn = n2[2a + (n – 1)d], we get
S10 = 102 [2 × 2 + (10 – 1)5]
= 5(4 + 9 × 5)
= 5(4 + 45) = 5 × 49 = 245

(ii) Let a be the first term and d be the common difference of the given AP. Then, we have:
a = – 37
and d = – 33 – (- 37) = – 33 + 37 = 4
We have to find the sum of 12 terms of the given AP.
Putting a = – 37, d = 4 and n = 12 in
Sn = n2[2a + (n – 1)d] we get,
S12 = 122[2 × (- 37) + (12 – 1)4]
= 6(- 74 + 11 × 4)
= 6(- 74 + 44) = 6 × (- 30) = – 180

(iii) Let a be the first term and d be the common difference of the given AP. Then, we have:
a = 0.6 and d = 1.7 – 0.6 = 1.1
We have to find the sum of 100 terms of the given AP.
Putting a = 0.6, d = 1.1 and n = 100 in
Sn = n2[2a + (n – 1)d], we get
S100 = 1002[2 × 0.6 + (100 – 1)1.1]
= 50(1.2 + 99 × 1.1)
= 50(1.2 + 108.9)
= 50 × 110.1
= 5505

(iv) Let a be the first term and d be the common difference of the given AP. Then, we have:
a = 115
and d = 112 – 115 = 5−460 = 160
We have to find the sum of 11 terms of the given AP.
Putting a = 115, d = 160 and n = 11 in

Question 2.
Find the sums given below:
(i) 7 + 1012 + 14 + ………… + 84
(ii) 34 + 32 + 30 + ………… + 10
(iii) – 5 + (- 8) + (- 11) + ………… + (- 230)
Solution:
(i) Here, the last term is given . We will first have to find the number of terms.

(ii) Here, the last term is given. We will first have to find the number of terms.
a = 34, d = 32 – 34 = – 2, l = an = 10
∴ 10 = a + (n – 1)d
or 10 = 34 + (n – 1)(- 2)
or (- 2)(n – 1) = 10 – 34 or (- 2)(n – 1) = – 24
or n – 1 = 12 or n = 12 + 1 = 13
Using Sn = 132(34 + 10) = 132 × 44
= 13 × 22 = 286

(iii) Here, the last term is given. We will first have to find the number of terms.
a = – 5, d = – 8 – (- 5) = – 8 + 5 = – 3,
l = an = – 230
∴ – 230 = a + (n – 1)d
or – 230 = – 5 + (n – 1)(- 3)
or (- 3)(n – 1) = – 230 + 5
or (- 3)(n – 1) = – 225
or n – 1 = −225−3 or n – 1 = 75
or n = 75 + 1 = 76
Using Sn = n2(a + l), we have:
S76 = 762 (- 5 – 230)
= 38 × (- 235) = – 8930

Question 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) We have:
a = 5, d = 3 and an = 50.
So, a + (n – 1)d = 50 or 5 + (n – 1)3 = 50
or 3(n – 1) = 50 – 5 or n – 1 = 453 = 15
or n = 15 + 1 = 16
Putting n = 16, a = 5 and l = an = 50 in Sn = n2 (a + l), we get
S16 = 162(5 + 50) = 8 × 55 = 440
Hence, n = 16 and S16 = 440.

(ii) We have:
a = 7 and a13 = 35.
Let d be the common difference of the given AP. Then,
a13 = 35 or a + 12d = 35
or 7 + 12d = 35 [∵ a = 7]
or 12d = 35 – 7 = 28 or d = 2812 = 73
Putting n = 13, a = 7 and l = a13 = 35 in Sn = n2(a + l), we get
S13 = 132(7 + 35) = 132 × 42
= 13 × 21 = 273
Hence, d = 73 and S13 = 273.

(iii) We have:
a12 = 37 and d = 3.
Let a be the first term of the given AP. Then,
a12 = 37 gives a + 11d = 37
So, a + 11(3) = 37 [∵ d = 3]
or a = 37 – 33 = 4
Putting n = 12, a = 4 and l = a12 = 37 in Sn = n2(a + l), we get
S12 = 122 (4 + 37) = 6 × 41 = 246
Hence, a = 4 and S12 = 246.

(iv) We have:
a3 = 15 and S10 = 125.
Let a be the first term and d be the common difference of the given AP. Then,
a3 = 15 and S10 = 125
So, a + 2d = 15 ………………… (1)
and 102 [2a + (10 – 1)d] = 125
or 5(2a + 9d) = 125
or 2a + 9d = 25 ……………… (2)
2 × (1) – (2) gives
2(a + 2d) – (2a + 9d) = 2 × 15 – 25
or 4d – 9d = 30 – 25
or – 5d = 5 or d = – 55 = – 1
Now, a10 = a + 9d = (a + 2d) + 7d
= 15 + 7(- 1) [using (1)]
= 15 – 7 = 8
Hence, d = – 1 and a10 = 8.

(v) We have:
d = 5 and S9 = 75.
Let a be the first term of the given AP. Then,
S9 = 75
So, 92 [2a + (9 – 1)5] = 75
or 92 (2a + 40) = 75
or 9a + 180 = 75
or 9a = 75 – 180
or 9a = – 105

(vi) We have:
a = 2, d = 8 and Sn = 90.
Sn = 90 gives
n2[2 × 2 + (n – 1)8] = 90
So, n2(4 + 8n – 8) = 90
or n2(8n – 4) = 90
or n(4n – 2) = 90
or 4n2 – 2n – 90 = 0

But n cannot be negative.
∴ n = 5
Now, an = a + (n – 1)d
or a5 = 2 + (5 – 1)8 = 2 + 32 = 34
Hence, n = 5 and an = 34.

(vii) We have:
a = 8, an = 62 and Sn = 210.
Let d be the common difference of the given AP.
Now, Sn = 210
So, n2(a + l) = 210
or n2(8 + 62) = 210 [∵ a = 8, an = 62]
or n2 × 70 = 210
or n = 210 × 270 = 3 × 2 = 6
and so an = 62 gives a6 = 62
So, a + 5d = 62
or 8 + 5d = 62 [∵ a = 8]
or 5d = 62 – 8 = 54 or d = 545
Hence, d = 545 and n = 6.

(viii) We have:
an = 4, d = 2 and Sn = – 14.
Let a be the first term of the given AP. Then,
an = 4 gives
a + (n – 1)2 = 4 [∵ d = 2]
or a = 4 – 2(n – 1) …………. (1)
and Sn = – 14 gives
n2(a + l) = – 14 [∵ l = an]
or n(a + 4) = – 28
or n[4 – 2(n – 1) + 41 = – 28
or n(4 – 2n + 2 + 4) = – 28
or n(- 2n + 10) = – 28
or n(- n + 5) = – 14
or – n2 + 5n = – 14
or n2 – 5n + 14 = 0
or (n – 7)(n + 2) = 0
i.e., n = 7 or – 2
But n cannot be negative.
Putting n = 7 in (1), we get
a = 4 – 2(7 – 1) = 4 – 2 × 6
= 4 – 12 = – 8
Hence, n = 7 and a = – 8.

(ix) We have:
a = 3, n – 8 and S = 192.
Let d be the common difference of the given AP.
Sn = n2[2a + (n – 1)d] gives
192 = 82 [2 × 3 + (8 – 1)d]
or 192 = 4(6 + 7d) or 48 = 6 + 7d
or 7d = 48 – 6 or 7d = 42
or d = 427 = 6
Hence, d = 6

(x) We have:
l = 28, S = 144 and n = 9.
Let a be the first term of the given AP.
S = 144 gives n2(a + l) = 144
or 92(a + 28) = 144 or a + 28 = 144 × 29
or a + 28 = 32 or a = 32 – 28 = 4
Hence, a = 4.

Question 4.
How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
Let the first term be a = 9 and common difference be d = 17 – 9 = 8.
Let the sum of n terms be 636. Then,
Sn = 636
So, n2[2a + (n – 1)d] = 636
or n2[2 × 9 + (n – 1)d] = 636
or n2(18 + 8n – 8) = 636
or n2(8n + 10) = 636
or n(4n + 5) = 636 or 4n2 + 5n – 636 = 0

But n cannot be negative.
∴ n = 12.
Thus, the sum of 12 terms is 636.

Question 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:
Let a be the first term and d be the common difference of the AP such that
a = 5, l = 45 and S = 400
∴ S = 400 gives n2(a + l) = 400
or n(5 + 45) = 400 × 2 or n(50) = 400 × 2
or n = 400×250 = 8 × 2 = 16
Also, l = 45
So, a + (n – 1)d = 45
or 5 + (16 – 1)d = 45
or 15d = 45 – 5 = 40 or d = 4015 = Hence, the number of terms is 16 and the common difference is 83.

Question 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:
Let a be the first term and d be the common difference. Let l be its last term. Then, a – 17, l = an = 350 and d = 9.
l = an = 350 gives
a + (n – 1)d = 350
or 17 + (n – 1)9 = 350
or 9(n – 1) = 350 – 17 = 333
or n – 1 = 3339 = 37 or n = 37 + 1 = 38
Putting a = 17, l = 350 and n = 38 in Sn = n2(a + l), we get
S38 = 382(17 + 350)
= 19 × 367 = 6973
Hence, there are 38 terms in the AP having their sum as 6973.

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Let a be the first term and d be the common difference of the given AP. Then,
d = 7 and a22 = 149
So, a + (22 – 1)d = 149 or a + 21 × 7 = 149
or a = 149 – 147 = 2
Putting n = 22, a – 2 and d = 7 in Sn = n2[2a + (n – 1)d], we get
S22 = 222[2 × 2 + (22 – 1)7]
= 11(4 + 21 × 7) = 11(4 + 147)
= 11 × 151 = 1661
Hence, the sum of first 22 terms of the AP is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Let a be the first term and d be the common difference of the given AP. Then,
a2 = 14 and a3 = 18
So, a + d = 14 and a + 2d = 18
Solving these equations, we get
d = 4 and a = 10 Putting a = 10,
d = 4 and n = 51 in

= 51 × 110 = 5610

Question 9.
If the sum of 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of n terms.
Solution:
Let a be the first term and d be the common difference of the given AP. Then,
S7 = 49 and S17 = 289

Solving these two equations, we get
d = 2 and a = 1

Question 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:
an = 3 + 4n
Substituting n = 1, 2, 3, 4, …………. n, we get
The sequence 7, 11, 15, 19, …, (3 + 4n), which is an AP with common difference 4.
Putting a = 7, d = 4 and n = 15 in Sn = n2[2a + (n – 1)d], we get

= 15 × 35 = 525

(ii) We have:
an = 9 – 5n
Substituting n = 1, 2, 3, 4, n, we get the sequence 4, – 1, – 6, – 11, …, (9 – 5n), which is an AP with common difference – 5.
Putting a = 4, d = – 5 and n = 15 in

Question 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:
According to the question,
Sn = 4n – n2.
∴ S1 = 4 × 1 – 12 = 4 – 1 = 3
Thus, First term = 3
Now, sum of first two terms = S2 = 4 × 2 – 22
= 8 – 4 = 4
∴ Second term = S2 – S1 = 4 – 3 = 1
S3 = 4 × 3 – 32 = 12 – 9 = 3
∴ Third term = S3 – S2 = 3 – 4 = – 1
S3 = 4 × 3 – 32
= 36 – 81
= – 45
and S10 = 4 × 10 – 102
= 40 – 100 = – 60
∴ Tenth term = S10 – S9
= – 60 – (- 45)
= – 60 + 45 = – 15
Also, Sn = 4n – n2
and Sn-1 = 4(n – 1) – (n – 1)2
= 4n – 4 – n2 + 2n – 1
= – n2 + 6n – 5
∴ nth term = Sn – Sn-1
= 4n – n2 – (- n2 + 6n – 5)
= 4n – n2 + n2 – 6n + 5
= 5 – 2n

Question 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, ….. Clearly, it is an AP with first term a = 6 and common difference d = 6. We want to find S40.
∴ S40 = 402 [2 × 6 + (40 – 1)6]
= 20(12 + 39 × 6)
= 20(12 + 234) = 20 × 246 = 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are 8 × 1, 8 × 2, 8 × 3, …………., 8 × 15
i.e., 8, 16, 24, …, 120, which is an AP.
∴ Sum of 1st 15 multiples of 8 = 152(8 + 120)
[Sn = n2(a + l)]
= 152 × 128
= 15 × 64 = 960

Question 14.
Find the sum of all the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 are 1, 3, 5, …, 49. They form an AP and there are 25 terms.
∴ Their sum = 252 (1 + 49) = 252 × 50
= 25 × 25 = 625

Question 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:
Here, a = 200, d = 50 and n = 30.
∴ S = 302[2 × 200 + (30 – 1)50]
[∵ Sn = n2(2a + (n – 1)d]
= 15(400 + 29 × 50)
= 15(400 + 1450)
= 15 × 1850
= 27750
Hence, a delay of 30 days costs the contractor Rs 27750.

Question 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:
Let the respective prizes be a + 60, a + 40, a + 20, a, a – 20, a – 40, a – 60
∴ The sum of the prizes is
a + 60 + a + 40 + a + 20 + a + a – 20 + a – 40 + a – 60 = 700
or 7a = 700 or a = 7007 = 100
∴ The seven prizes are 100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60, i.e., 160, 140, 120, 100, 80, 60, 40 (in Rs).

Question 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:
Since there are three sections of each class, so the number of trees planted by class I, class II, class III, …, class XII are 1 × 3, 2 × 3, 3 × 3, ………….., 12 × 3 respectively.
i.e., 3, 6, 9, …, 36. Clearly, it forms an AP.
The sum of the number of the trees planted by these classes
= 122(3 + 36) = 6 × 39 = 234

Question 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, … as shown in fig. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 227)

Solution:
Length of a semi-circumference = πr, where r is the radius of the circle.
∴ Length of spiral made up of thirteen consecutive semicircles
= (π × 0.5 + π × 1.0 + π × 1.5 + π × 2.0 + … + π × 6.5) cm
= π × 0.5(1 + 2 + 3 + … + 13) cm
= π × 0.5 × 132(2 × 1 + [13 – 1) × 1] cm
= 227 × 510 × 132 × 14 cm = 143 cm

Question 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 (see figure).
In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:
Clearly, logs stacked in each row form a sequence 20, 19, 18, 17, …. It is an AP with a = 20 and d = 19 – 20 = – 1.
Let Sn = 200. Then,
n2[2 × 20 + (n – 1)(- 1)] = 200
or n(40 – n + 1) = 400
or n2 – 41n + 400 = 0
or (n – 16)(n – 25) = 0
i.e., n = 16 or 25
Here, the common difference is negative.
The terms go on diminishing and 21st term becomes zero. All terms after 21st term are negative.
These negative terms when added to positive terms from 17th term to 20th term, cancel out each other and the sum remains the same.
Thus, n = 25 is not valid for this problem. So, we take n = 16
Thus, 200 logs are placed in 16 rows.
Number of logs in the 16th row = a16
= a + 15d
= 20 + 15(- 1)
= 20 – 15 = 5

Question 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 (see figute below).

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 first potato, second potato, third potato, fourth potato, …………,
the distances (in metres) run by the competitor are
2 × 5; 2 × (5 + 3), 2 × (5 + 3 + 3), 2 × (5 + 3 + 3 + 3), …
i.e., 10, 16, 22, 28, …
which is in AP with a = 10 and d = 16 – 10 = 6.
The sum of first ten terms,
S10 = 102 [(2 × 10 + (10 – 1) × 6)]
= 5(20 + 54)
= 5 × 74
= 370
∴ The total distance the competitor has to run is 370 m.

BSEB Bihar Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.
Which term of the AP : 121, 117, 113, …, is its first negative term?
Solution:
121, 117, 113, …
a = 121, d = 117 – 121 = – 4
an = a + (n – 1 )d
= 121 + in – 1) x (- 4)
= 121 – 4n + 4 = 125 – 4n
For the first negative term.
an < 0 i.e., 125 – 4n < 0
or 125 < 4n or 1254 < n
or 3114 < n n is an integer and n >3114
∴ The first negative term is 32nd term.

Question 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:
Let the AP be a – 4d, a – 3d, a – 2d, a – d, a, a + d, a + 2d, a + 3d, …
Then, a3 = a – 2d, a7 = a – 2d
So, a3+a7 = a – 2d + a – 2d = 6
or 2a = 6 or a – 3
Also, (a – 2d)(a + 2d) = 8
So, a²-4d² = 8 or 4d² = a² – 8
or 4d² = (3)² – 8 = 9 – 8 = 1
or d² = 14 or d = ± 12
Taking d = 12

Question 3.
A ladder has rungs 25 cm apart (see figure). The rungs decrease uniformly in length from 45 cm, at the bottom to 25 cm at the top. If the top and the
bottom rungs are 212 m apart, what is the length of the wood required for the rungs?

Solution:
Number of rungs, n = 212 m25 cm = 250 cm25 cm = 10
So, there are 10 rungs.
The length of the wood required for rungs
= sum of 10 rungs
= 102 [25 + 45] cm
= 5 x 70 cm
= 350 cm.

Question 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:

According to the question,

∵ x is a counting number, so taking positive square root, x = 7 x 5 = 35.

Question 5.
A small terrace at a football ground comprises 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 14 m and a tread of 12 (see figure). Calculate the total volume of concrete required to build the terrace.

Solution:
Volume of concrete required to build the first step, second step, third step, … (in m³) are
14 x 12 x 50, ( 2 x 14 ) x 12 x 50, (3 x 14) x 12 x 50, ….
i.e., 508 x 2 x 508, 3 x 508, ….
∴ Total volume of concrete required
= [508 + 2 x 508 + 3 x 508 + … ] m³
= 508[1 + 2 + 3 + …] m³
= 508 x 508 [2 x 1 + (15 – 1) x 1] m³ [∵ n = 15]
= 508 x 508 x 16 = 750 m³