Exercise 5.1

1. In which of the following situations, does the list of numbers involved make as 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.
Answer:
It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39
Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Answer:
Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 – 1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)² , (3V/4)³…
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(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.
Answer:
Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Answer:

We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be 

Therefore, after every year, our money will be

 

It is not an arithmetic progression because (2) − (1) ≠ (3) − (2)

(Difference between consecutive terms is not equal)

Therefore, it is not an Arithmetic Progression.

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

Answer:
(i) a = 10, d = 10
Let the series be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = 10
a₂ = a₁ + d = 10 + 10 = 20
a₃ = a₂ + d = 20 + 10 = 30
a₄ = a₃ + d = 30 + 10 = 40
a₅ = a₄ + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = – 2, d = 0
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = -2
a₂ = a₁ + d = – 2 + 0 = – 2
a₃ = a₂ + d = – 2 + 0 = – 2
a₄ = a₃ + d = – 2 + 0 = – 2
Therefore, the series will be – 2, – 2, – 2, – 2 …
First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii) a = 4, d = – 3
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = 4
a₂ = a₁ + d = 4 – 3 = 1
a₃ = a₂ + d = 1 – 3 = – 2
a₄ = a₃ + d = – 2 – 3 = – 5
Therefore, the series will be 4, 1, – 2 – 5 …
First four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) a = – 1, d = 1/2
Let the series be a₁, a₂, a₃, a₄ …a₁ = a = -1
a₂ = a₁ + d = -1 + 1/2 = -1/2
a₃ = a₂ + d = -1/2 + 1/2 = 0
a₄ = a₃ + d = 0 + 1/2 = 1/2
Clearly, the series will be-1, -1/2, 0, 1/2
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = – 1.25, d = – 0.25
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = – 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 = – 2.00
Clearly, the series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
First four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
(ii) -5, – 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 …

Answer:
(i) 3, 1, – 1, – 3 …
Here, first term, a = 3
Common difference, d = Second term – First term
= 1 – 3 = – 2

(ii) – 5, – 1, 3, 7 …
Here, first term, a = – 5
Common difference, d = Second term – First term
= ( – 1) – ( – 5) = – 1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3 ….
Here, first term, a = 1/3
Common difference, d = Second term – First term
= 5/3 – 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term, a = 0.6
Common difference, d = Second term – First term
= 1.7 – 0.6
= 1.1

4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a², a³, a⁴ …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 1², 3², 5², 7² …
(xv) 1², 5², 7², 7³ …

Answer:
(i) 2, 4, 8, 16 …
Here,
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
⇒ a_n+1 – an is not the same every time.
Therefore, the given numbers are forming an A.P.

(ii) 2, 5/2, 3, 7/2 ….
Here,
a₂ – a₁ = 5/2 – 2 = 1/2
a₃ – a₂ = 3 – 5/2 = 1/2
a₄ – a₃ = 7/2 – 3 = 1/2
⇒ a_n+1 – a_n is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
a₅ = 7/2 + 1/2 = 4
a₆ = 4 + 1/2 = 9/2
a₇ = 9/2 + 1/2 = 5

(iii) -1.2, – 3.2, -5.2, -7.2 …
Here,
a₂ – a₁ = ( -3.2) – ( -1.2) = -2
a₃ – a₂ = ( -5.2) – ( -3.2) = -2
a₄ – a₃ = ( -7.2) – ( -5.2) = -2
⇒ a_n+1 – a_n is same every time.
Therefore, d = -2 and the given numbers are in A.P.
Three more terms are
a₅ = – 7.2 – 2 = – 9.2
a₆ = – 9.2 – 2 = – 11.2
a₇ = – 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
⇒ a_n+1 – a_n is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a₅ = 2 + 4 = 6
a₆ = 6 + 4 = 10
a₇ = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a₂ – a₁ = 3 + √2 – 3 = √2
a₃ – a₂ = (3 + 2√2) – (3 + √2) = √2
a₄ – a₃ = (3 + 3√2) – (3 + 2√2) = √2
⇒ a_n+1 – a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a₅ = (3 + √2) + √2 = 3 + 4√2
a₆ = (3 + 4√2) + √2 = 3 + 5√2
a₇ = (3 + 5√2) + √2 = 3 + 6√2

(vi) 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
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(vii) 0, -4, -8, -12 …
Here,
a₂ – a₁ = (-4) – 0 = -4
a₃ – a₂ = (-8) – (-4) = -4
a₄ – a₃ = (-12) – (-8) = -4
⇒ a_n+1 – a_n is same every time.
Therefore, d = -4 and the given numbers are in A.P.
Three more terms are
a₅ = -12 – 4 = -16
a₆ = -16 – 4 = -20
a₇ = -20 – 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….
Here,
a₂ – a₁ = (-1/2) – (-1/2) = 0
a₃ – a₂ = (-1/2) – (-1/2) = 0
a₄ – a₃ = (-1/2) – (-1/2) = 0
⇒ a_n+1 – a_n is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a₅ = (-1/2) – 0 = -1/2
a₆ = (-1/2) – 0 = -1/2
a₇ = (-1/2) – 0 = -1/2

(ix) 1, 3, 9, 27 …
Here,
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
a₄ – a₃ = 27 – 9 = 18
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(x) a, 2a, 3a, 4a …
Here,
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
⇒ a_n+1 – a_n is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a₅ = 4a + a = 5a
a₆ = 5a + a = 6a
a₇ = 6a + a = 7a

(xi) a, a₂, a₃, a₄ …
Here,
a₂ – a₁ = a^₂ – a = (a – 1)
a₃ – a₂ = a^₃ – a_² = a₂ (a – 1)
a₄ – a₃ = a^₄ – a^₃ = a₃(a – 1)
⇒ an+1 – an is not the same every time.
Therefore, the given numbers are forming an A.P.

(xii) √2, √8, √18, √32 …
Here,
a₂ – a₁ = √8 – √2  = 2√2 – √2 = √2
a₃ – a₂ = √18 – √8 = 3√2 – 2√2 = √2
a₄ – a₃ = 4√2 – 3√2 = √2
⇒ a_n+1 – a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a₅ = √32  + √2 = 4√2 + √2 = 5√2 = √50
a₆ = 5√2 +√2 = 6√2 = √72
a₇ = 6√2 + √2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …
Here,
a₂ – a₁ = √6 – √3 = √3 × 2 -√3 = √3(√2 – 1)
a₃ – a₂ = √9 – √6 = 3 – √6 = √3(√3 – √2)
a₄ – a₃ = √12 – √9 = 2√3 – √3 × 3 = √3(2 – √3)
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(xiv) 1², 3², 5², 7² …
Or, 1, 9, 25, 49 …..
Here,
a₂ − a₁ = 9 − 1 = 8
a₃ − a₂ = 25 − 9 = 16
a₄ − a₃ = 49 − 25 = 24
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(xv) 12, 5², 7², 73 …
Or 1, 25, 49, 73 …
Here,
a₂ − a₁ = 25 − 1 = 24
a₃ − a₂ = 49 − 25 = 24
a₄ − a₃ = 73 − 49 = 24
i.e., a_k+1 − a_k is same every time.
⇒ a_n+1 – a_n is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a₅ = 73+ 24 = 97
a₆ = 97 + 24 = 121
a₇ = 121 + 24 = 145

Exercise 5.2

1. Find the missing variable from a, d, n and an, where a is the first term, d is the common difference and an is the nth term of AP.

(i) a = 7, d = 3, n = 8

(ii) a = –18, n = 10, a_n=0

(iii) d = –3, n = 18, a_n=-5

(iv) a = –18.9, d = 2.5, a_n=-3.6

(v) a = 3.5, d = 0, n = 105

Answer:

(i) a = 7, d = 3, n = 8

We know that,

= 7 + (8 − 1) 3

= 7 + (7) 3

= 7 + 21 = 28

(ii) a = –18, n= 10, 

We need to find d here.

Using formula 

Putting values of a,n and ,

0 = –18 + (10 – 1) d

⇒ 0 = −18 + 9d

⇒ 18 = 9d ⇒ d = 2

(iii) d = –3, n = 18, 

We need to find a here.

Using formula 

Putting values of d, ,

–5 = a + (18 – 1) (–3)

⇒ −5 = a + (17) (−3)

⇒ −5 = a – 51 ⇒ a = 46

(iv) a = –18.9, d = 2.5, 

We need to find n here.

Using formula 

Putting values of d, a,and ,

3.6 = –18.9 + (n – 1) (2.5)

⇒ 3.6 = −18.9 + 2.5n − 2.5

⇒ 2.5n = 25 ⇒ n = 10

(v) a = 3.5, d = 0, n = 105

We need to find here.

Using formula 

Putting values of d, n and a,

an = 3.5 + (105 − 1) (0)

⇒

2. Choose the correct choice in the following and justify:

(i) 30^th term of the AP: 10, 7, 4… is

(A) 97

(B) 77

(C) –77

(D) –87

(ii) 11^th term of the AP: −3, −12, 2… is

(A) 28

(B) 22

(C) –38

(D) 

Answer:

(i) 10, 7, 4…

First term = a = 10, Common difference = d = 7 – 10 = 4 – 7 = –3

And n = 30{Because, we need to find 30^th term}

⇒ = 10 + (30 − 1) (−3) = 10 – 87 = −77

Therefore, the answer is (C).

(ii) −3, −½, 2…

First term = a = –3, Common difference = d = − − (−3) = 

And n = 11 (Because, we need to find 11th term)

= −3 + (11 – 1) = −3 + 25 = 22

Therefore 11^th term is 22 which means answer is (B).

3. In the following AP’s find the missing terms:

(i) 2, __ , 26

(ii) __, 13, __, 3

(iii) 5, __, __, 

(iv) –4. __, __, __, __, 6

(v) __, 38, __, __, __, –22

Answer:

(i) For this A.P.,
a = 2
a₃ = 26
We know that, an = a + (n − 1) d
a₃ = 2 + (3 – 1) d
26 = 2 + 2d
24 = 2d
d = 12
a₂ = 2 + (2 – 1) 12
= 14
Therefore, 14 is the missing term.

(ii) For this A.P.,
a₂ = 13 and
a₄ = 3
We know that, an = a + (n − 1) d
a₂ = a + (2 – 1) d
13 = a + d … (i)
a₄ = a + (4 – 1) d
3 = a + 3d … (ii)
On subtracting (i) from (ii), we get
– 10 = 2d
d = – 5
From equation (i), we get
13 = a + (-5)
a = 18
a₃ = 18 + (3 – 1) (-5)
= 18 + 2 (-5) = 18 – 10 = 8
Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,
a = 5 and
a₄ = 19/2
We know that, an = a + (n − 1) d
a₄ = a + (4 – 1) d
19/2 = 5 + 3d
19/2 – 5 = 3d3d = 9/2
d = 3/2
a₂ = a + (2 – 1) d
a₂ = 5 + 3/2
a₂ = 13/2
a₃ = a + (3 – 1) d
a₃ = 5 + 2×3/2
a₃ = 8
Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P.,
a = −4 and
a₆ = 6
We know that,
a_n = a + (n − 1) d
a₆ = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a₂ = a + d = − 4 + 2 = −2
a₃ = a + 2d = − 4 + 2 (2) = 0
a₄ = a + 3d = − 4 + 3 (2) = 2
a₅ = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v)For this A.P.,
a₂ = 38
a₆ = −22
We know that
a_n = a + (n − 1) d
a₂ = a + (2 − 1) d
38 = a + d … (i)
a₆ = a + (6 − 1) d
−22 = a + 5d … (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a₂ − d = 38 − (−15) = 53
a₃ = a + 2d = 53 + 2 (−15) = 23
a₄ = a + 3d = 53 + 3 (−15) = 8
a₅ = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

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

Answer:
3, 8, 13, 18, …
For this A.P.,
a = 3
d = a₂ − a1 = 8 − 3 = 5
Let n^th term of this A.P. be 78.
an = a + (n − 1) d
78 = 3 + (n − 1) 5
75 = (n − 1) 5
(n − 1) = 15
n = 16
Hence, 16^th term of this A.P. is 78.

5. Find the number of terms in each of the following APs:

(i) 7, 13, 19…., 205

(ii) 18, , 13…, −47

Answer:

(i) For this A.P.,
a = 7
d = a₂ − a₁ = 13 − 7 = 6
Let there are n terms in this A.P.
a_n = 205
We know that
a_n = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms in it.

(ii) 18, , 13 …, −47

First term = a =18, Common difference = d =

 

Using formula , to find nth term of arithmetic progression,

−47 = 18 + (n − 1) 

= 

Therefore, there are 27 terms in the given arithmetic progression.

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

Answer:
For this A.P.,
a = 11
d = a₂ − a₁ = 8 − 11 = −3
Let −150 be the nth term of this A.P.
We know that,
an = a + (n − 1) d
-150 = 11 + (n – 1)(-3)
-150 = 11 – 3n + 3
-164 = -3_n
n = 164/3
Clearly, n is not an integer.
Therefore, – 150 is not a term of this A.P.

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

Answer:
Given that,
a₁₁ = 38
a₁₆ = 73
We know that,
a_n = a + (n − 1) d
a₁₁ = a + (11 − 1) d
38 = a + 10d … (i) Similarly,
a₁₆ = a + (16 − 1) d
73 = a + 15d … (ii)
On subtracting (i) from (ii), we get
35 = 5d
d = 7
From equation (i),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a₃₁ = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31^st term is 178.

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

Answer:
Given that,
a₃ = 12
a₅₀ = 106
We know that,
a_n = a + (n − 1) d
a₃ = a + (3 − 1) d
12 = a + 2d … (i)
Similarly, a₅₀ = a + (50 − 1) d
106 = a + 49d … (ii)
On subtracting (i) from (ii), we get
94 = 47d
d = 2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a₂₉ = a + (29 − 1) d
a₂₉ = 8 + (28)²
a₂₉ = 8 + 56 = 64
Therefore, 29^th term is 64.

9. If the 3^rd and the 9^th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

Answer:
Given that,
a₃ = 4
a₉ = −8
We know that,
a_n = a + (n − 1) d
a₃ = a + (3 − 1) d
4 = a + 2d … (i)
a₉ = a + (9 − 1) d
−8 = a + 8d … (ii)
On subtracting equation (i) from (ii), we get,
−12 = 6d
d = −2
From equation (i), we get,
4 = a + 2 (−2)
4 = a − 4
a = 8
Let nth term of this A.P. be zero.
a_n = a + (n − 1) d
0 = 8 + (n − 1) (−2)
0 = 8 − 2n + 2
2_n = 10
n = 5
Hence, 5th term of this A.P. is 0.

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

Answer:
We know that,
For an A.P., an = a + (n − 1) d
a₁₇ = a + (17 − 1) d
a₁₇ = a + 16d
Similarly, a₁₀ = a + 9d
It is given that
a₁₇ − a₁₀ = 7
(a + 16d) − (a + 9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.

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

Answer:
Given A.P. is 3, 15, 27, 39, …
a = 3
d = a₂ − a₁ = 15 − 3 = 12
a₅₄ = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let n^th term be 771.
an = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65th term was 132 more than 54th term.

12. Two APs have the same common difference. The difference between their 100^th term is 100, what is the difference between their 1000^th terms?

Answer:
Let the first term of these A.P.s be a₁ and a₂ respectively and the common difference of these A.P.s be d.
For first A.P.,
a₁₀₀ = a₁ + (100 − 1) d
= a₁ + 99d
a₁₀₀₀ = a₁ + (1000 − 1) d
a₁₀₀₀ = a₁ + 999d
For second A.P.,
a₁₀₀ = a₂ + (100 − 1) d
= a2 + 99d
a₁₀₀₀ = a₂ + (1000 − 1) d
= a₂ + 999d
Given that, difference between
100^th term of these A.P.s = 100
Therefore, (a₁ + 99d) − (a₂ + 99d) = 100
a₁ − a₂ = 100 … (i)
Difference between 1000^th terms of these A.P.s
(a₁ + 999d) − (a₂ + 999d) = a₁ − a₂
From equation (i),
This difference, a₁ − a₂ = 100
Hence, the difference between 1000^th terms of these A.P. will be 100.

13. How many three digit numbers are divisible by 7?

Answer:
First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
a_n = 994
n = ?
a_n = a + (n − 1) d
994 = 105 + (n − 1) 7
889 = (n − 1) 7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

Answer:
First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be the nth term of this A.P.
a = 12
d = 4
a_n = 248
a_n = a + (n – 1) d
248 = 12 + (n – 1) × 4
236/4 = n – 1
59  = n – 1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.

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

Answer:
63, 65, 67, …
a = 63
d = a₂ − a1 = 65 − 63 = 2
n^th term of this A.P. = a_n = a + (n − 1) d
a_n= 63 + (n − 1) 2 = 63 + 2_n − 2
a_n = 61 + 2_n … (i)
3, 10, 17, …
a = 3
d = a₂ − a₁ = 10 − 3 = 7
n^th term of this A.P. = 3 + (n − 1) 7
a_n = 3 + 7_n − 7
a_n = 7_n − 4 … (ii)
It is given that, n^th term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2_n = 7_n − 4
61 + 4 = 5_n
5_n = 65
n = 13
Therefore, 13^th terms of both these A.P.s are equal to each other.

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

Answer:
a₃ = 16
a + (3 − 1) d = 16
a + 2d = 16 … (i)
a₇ − a₅ = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …

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

Answer:
Given A.P. is
3, 8, 13, …, 253
Common difference for this A.P. is 5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 5
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a₂₀ = a + (20 − 1) d
a₂₀ = 253 + (19) (−5)
a₂₀ = 253 − 95
a = 158
Therefore, 20^th term from the last term is 158.

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

Answer:
We know that,
a_n = a + (n − 1) d
a₄ = a + (4 − 1) d
a₄ = a + 3d
Similarly,
a₈ = a + 7d
a₆ = a + 5d
a₁₀ = a + 9d
Given that, a₄ + a₈ = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 … (i)
a₆ + a₁₀ = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 … (ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5
From equation (i), we get
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a₂ = a + d = − 13 + 5 = −8
a₃ = a₂ + d = − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.

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?

Answer:
It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Here, a = 5000
d = 200
Let after nth year, his salary be Rs 7000.
Therefore, a_n = a + (n − 1) d
7000 = 5000 + (n − 1) 200
200(n − 1) = 2000
(n − 1) = 10
n = 11
Therefore, in 11^th year, his salary will be Rs 7000.

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

Answer:
Given that,
a = 5
d = 1.75
a_n = 20.75
n = ?
a_n = a + (n − 1) d
20.75 = 5 + (n – 1) × 1.75
15.75 = (n – 1) × 1.75
(n – 1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n – 1 = 9
n = 10
Hence, n is 10.

Exercise 5.3

1. Find the sum of the following AP’s.

(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

Answer:

(i) 2, 7, 12… to 10 terms

Here First term = a = 2, Common difference = d = 7 – 2 = 5 and n = 10

Applying formula,to find sum of n terms of AP,

(ii) –37, –33, –29… to 12 terms

Here First term = a = –37, Common difference = d = –33 – (–37) = 4

And n = 12

Applying formula,to find sum of n terms of AP,

(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,to find sum of n terms of AP,

(iv) to 11 terms

Here First tern =a = 1/15Common difference = 

Applying formula,to find sum of n terms of AP,

2. Find the sums given below
(i) 7 + 21/2+ 14 + ……………… +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Answer:

(i)

Here First term = a = 7, Common difference = d =

And Last term = l = 84

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula , to find n^th term of arithmetic progression,

[7 + (n − 1) (3.5)] = 84

⇒ 7 + (3.5) n − 3.5 = 84

⇒ 3.5n = 84 + 3.5 – 7

⇒ 3.5n = 80.5

⇒ n = 23

Therefore, there are 23 terms in the given AP.

It means n = 23.

Applying formula, to find sum of n terms of AP,

(ii) 34 + 32 + 30 + … + 10

Here First term = a = 34, Common difference = d = 32 – 34 = –2

And Last term = l = 10

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula , to find nth term of arithmetic progression,

[34 + (n − 1) (−2)] = 10

⇒ 34 – 2n + 2 = 10

⇒ −2n = −26⇒ n = 13

Therefore, there are 13 terms in the given AP.

It means n = 13.

Applying formula, to find sum of n terms of AP,

(iii) −5 + (−8) + (−11) + … + (−230)

Here First term = a = –5, Common difference = d = –8 – (–5) = –8 + 5 = –3

And Last term = l = −230

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula , to find nth term of arithmetic progression,

[−5 + (n − 1) (−3)] = −230

⇒ −5 − 3n + 3 = −230

⇒ −3n = −228 ⇒ n = 76

Therefore, there are 76 terms in the given AP.

It means n = 76.

Applying formula, to find sum of n terms of AP,

3. In an AP
(i) Given a = 5, d = 3, a_n = 50, find n and S_n.
(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, S_n = 90, find n and an.
(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) Given an = 4, d = 2, S_n = − 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.

Answer:

(i) Given a = 5, d = 3, a_n = 50,find n and S_n.

Using formula a_n = a + (n-1)d, to find nth term of arithmetic progression,

⇒ 50 = 5 + (n − 1) (3)

⇒ 50 = 5 + 3n − 3

⇒ 48 = 3n⇒ n = 16

Applying formula,to find sum of n terms of AP,

Therefore, n = 16 and S_n  = 440

(ii) Given a = 7, a₁₃ = 35, find d and S₁₃..

Using formula a_n = a + (n-1)d, to find n^th term of arithmetic progression,

a_n = a + (n-1)d

a₁₃= 7 + (13 − 1) (d)

⇒ 35 = 7 + 12d

⇒ 28 = 12d⇒ d = 28/12 = 7/3

Applying formula,to find sum of n terms of AP,

Therefore, d = 7/3 and S₁₃ = 273

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂..

Using formula a_n = a + (n-1)d, to find nth term of arithmetic progression,

a_n = a + (n-1)d

a₁₂ = a + (12 − 1) 3

⇒ 37 = a + 33 ⇒ a = 4

Applying formula,to find sum of n terms of AP,

Therefore, a = 4 and S12 = 246

(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀..

Using formula a_n = a + (n-1)d, to find n^th term of arithmetic progression,

a_n = a + (n-1)d

a₃ = a + (3 − 1) (d)

⇒ 15 = a + 2d

⇒ a = 15 − 2d… (1)

Applying formula,to find sum of n terms of AP,

⇒ 125 = 5 (2a + 9d) = 10a + 45d

Putting (1) in the above equation,

125 = 5 [2 (15 − 2d) + 9d] = 5 (30 − 4d + 9d)

⇒ 125 = 150 + 25d

⇒ 125 – 150 = 25d

⇒ −25 = 25d⇒ d = −1

Using formula a_n = a + (n-1)d, to find nth term of arithmetic progression,

a_n = a + (n-1)d

a₁₀= a + (10 − 1) d

Putting value of d and equation (1) in the above equation,

a₁₀= 15 − 2d + 9d = 15 + 7d

= 15 + 7 (−1) = 15 – 7 = 8

Therefore, d = −1 and a₁₀= 8

(v) Given d = 5,S₉ = 75, find a and a₉..

Applying formula,to find sum of n terms of AP,

⇒ 150 = 18a + 360

⇒ −210 = 18a

⇒ a = 

Using formula a_n = a + (n-1)d, to find n^th term of arithmetic progression,

a₉ = + (9 − 1) (5)

a₉ =

Therefore, a = and a9 = 

(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.

Applying formula,to find sum of n terms of AP,

⇒ 2n (n − 5) + 9 (n − 5) = 0

⇒ (n − 5) (2n + 9) = 0

⇒ n = 5,−9/2

We discard negative value of n because here n cannot be in negative or fraction.

The value of n must be a positive integer.

Therefore, n = 5

Using formula a_n = a + (n-1)d ,to find nth term of arithmetic progression,

a₅ = 2 + (5 − 1) (8) = 2 + 32 = 34

Therefore, n = 5 and a5 = 34

(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.

Using formula a_n = a + (n-1)d, to find nth term of arithmetic progression,

62 = 8 + (n − 1) (d) = 8 + nd – d

⇒ 62 = 8 + nd − d

⇒ nd – d = 54

⇒ nd = 54 + d… (1)

Applying formula,to find sum of n terms of AP,

⇒ n = 6

Putting value of n in equation (1),

6d = 54 + d ⇒ d = 

Therefore, n = 6 and d =

(viii) Given Given a_n = 4, d = 2, S_n = − 14, find n and a

Using formula a_n = a + (n-1)d , to find n^th term of arithmetic progression,

4 = a + (n − 1) (2) = a + 2n − 2

⇒ 4 = a + 2n – 2

⇒ 6 = a + 2n

⇒ a = 6 − 2n… (1)

Applying formula,to find sum of n terms of AP,

⇒ (n + 2) (n − 7) = 0

⇒ n = −2, 7

Here, we cannot have negative value of n.

Therefore, we discard negative value of n which means n = 7.

Putting 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.

Using formula, a_n = a + (n-1)d to find sum of n terms of AP, we get

192 = [6 + (8 − 1) d] = 4 (6 + 7d)

⇒ 192 = 24 + 28d

⇒ 168 = 28d ⇒ d = 6

(x) Given l = 28, S = 144, and there are total of 9 terms. Find a.

Applying formula, , to find sum of n terms, we get

144 = [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?

Answer:

First term = a = 9, Common difference = d = 17 – 9 = 8, Sn = 636

Applying formula,to find sum of n terms of AP, we get

636 =[18 + (n − 1) (8)]

⇒ 1272 = n (18 + 8n − 8)

⇒

We discard negative value of n here because n cannot be in negative, n can only be a positive integer.

Therefore, n = 12

Therefore, 12 terms of the given sequence make sum equal to 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.

Answer.

First term = a = 5, Last term = l = 45, 

Applying formula, to find sum of n terms of AP, we get

⇒ n = 16

Applying formula,to find sum of n terms of AP and putting value of n, we get

We get,

45 = 5 + (16 – 1)d

45 = 5 + (15)d

45 – 5 = 15d

 

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?

Answer:

First term = a = 17, Last term = l = 350 and Common difference = d = 9

Using formula , to find nth term of arithmetic progression, we get

350 = 17 + (n − 1) (9)

⇒ 350 = 17 + 9n − 9

⇒ 342 = 9n ⇒ n = 38

Applying formula,to find sum of n terms of AP and putting value of n, we get

 =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.

Answer:

It is given that 22nd term is equal to 149

Using formula a_n = a + (n-1)d, to find nth term of arithmetic progression, we get

a_n = a + (n-1)d

149 = a + (22 − 1) (7)

⇒ 149 = a + 147⇒ a = 2

Applying formula,to find sum of n terms of AP and putting value of a, we get

= 11 (4 + 147)

⇒ 11(151)= 1661

Therefore, sum of first 22 terms of AP is equal to 1661.

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

Answer:

It is given that second and third term of AP are 14 and 18 respectively.

Using formula a_n = a + (n-1)d, to find nth term of arithmetic progression, we get

14 = a + (2 − 1) d

⇒ 14 = a + d … (1)

And, 18 = a + (3 − 1) d

⇒ 18 = a + 2d … (2)

These are equations consisting of two variables.

Using equation (1), we get, a = 14 − d

Putting value of a in equation (2), we get

18 = 14 – d + 2d

⇒ d = 4

Therefore, common difference d = 4

Putting value of d in equation (1), we get

18 = a + 2 (4)

⇒ a = 10

Applying formula,to find sum of n terms of AP, we get

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.

Answer:

It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289.

Applying formula,to find sum of n terms of AP, we get

⇒ 98 = 7 (2a + 6d)

⇒ 7 = a + 3d ⇒ a = 7 − 3d … (1)

And, s₁₇= 289

⇒ 578 = 17 (2a + 16d)

⇒ 34 = 2a + 16d

⇒ 17 = a + 8d

Putting equation (1) in the above equation, we get

17 = 7 − 3d + 8d

⇒ 10 = 5d ⇒ d = 2

Putting value of d in equation (1), we get

a = 7 − 3d = 7 – 3 (2) = 7 – 6 = 1

Again applying formula,to find sum of n terms of AP, we get

10. Show that a₁, a₂ … , a_n , … form an AP where an is defined as below
(i) a_n = 3 + 4_n
(ii) a_n = 9 − 5_n
Also find the sum of the first 15 terms in each case.

Answer:

(i)

is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Therefore, sum of first 15 terms of AP is equal to 525.

(ii) 

is same every time. Therefore, this is an AP with common difference as -5 and first term as 4.

 

Therefore, sum of first 15 terms of AP is equal to –465.

11. If the sum of the first n terms of an AP is 4_n − 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 3^rd, the10^th and the nth terms.

Answer:

It is given that the sum of n terms of an AP is equal to 

It means 

So here, we can find the first term by substituting n = 1 ,

=4-1=3

Similarly, the sum of first two terms can be given by,

= 8 – 4

= 4

Now, as we know,

 

So,

 

 = 4 – 3

= 1

Now, using the same method we have to find the third, tenth and nth term of A.P.

So, for the third term,

 

=

= (12 – 9) – (8 – 4)

= 3 – 4

= – 1

Also, for the tenth term,

=

= (40 – 100) – (36 – 81)

= – 60 + 45

= 15

So, for the nth term,

= 

=

=

= 5 – 2n

Therefore,=

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

Answer:

The first positive integers that are divisible by 6 are 6, 12, 18, 24 … 40 terms.

Therefore, we want to find sum of 40 terms of sequence of the form:

6, 12, 18, 24 … 40 terms

Here, first term = a = 6 and Common difference = d = 12 – 6 = 6, n = 40

Applying formula, to find sum of n terms of AP, we get

a = 6

d = 6

S₄₀ = ?

=

= 20 [12 + (39) (6 )]

= 20 (12 + 234)

= 20 x 246

= 4920

13. Find the sum of the first 15 multiples of 8.

Answer:

The first 15 multiples of 8 are 8,16, 24, 32 … 15 times

First term = a = 8 and Common difference = d = 16 – 8 = 8, n = 15

Applying formula, to find sum of n terms of AP, we get

I  = a + (n-1) d

  = 8 + (15 – 1) 8

 = 120

Reruired Sum =

 =

Hence, the required sum is 960.

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

Answer:

The odd numbers between 0 and 50 are 1, 3, 5, 7 … 49

It is an arithmetic progression because the difference between consecutive terms is constant.

First term = a = 1, Common difference = 3 – 1 = 2, Last term = l = 49

We do not know how many odd numbers are present between 0 and 50.

Therefore, we need to find n first.

Using formula an = a + (n − 1) d, to find nth term of arithmetic progression, we get

49 = 1 + (n − 1) 2

⇒ 49 = 1 + 2n − 2

⇒ 50 = 2n ⇒ n = 25

Applying formula, to find sum of n terms of AP, we get

=

=

= 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?

Answer:

Penalty for first day = Rs 200, Penalty for second day = Rs 250

Penalty for third day = Rs 300

It is given that penalty for each succeeding day is Rs 50 more than the preceding day.

It makes it an arithmetic progression because the difference between consecutive terms is constant.

We want to know 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 form200, 250, 300, 350 … 30 terms

First term = a = 200, Common difference = d = 50, n = 30

Applying formula, to find sum of n terms of AP, we get

⇒

⇒ 15 (400 + 1450)

 = 15 x 1850

 = 27750

Therefore, penalty for 30 days is Rs. 27750.

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 term, find the value of each of the prizes.

Answer:

It is given that sum of seven cash prizes is equal to Rs 700.

And, each prize is R.s 20 less than its preceding term.

Let value of first prize = Rs. a

Let value of second prize =Rs (a − 20)

Let value of third prize = Rs (a − 40)

So, we have sequence of the form:

a, a − 20, a − 40, a – 60 …

It is an arithmetic progression because the difference between consecutive terms is constant.

First term = a, Common difference = d = (a – 20) – a = –20

n = 7 (Because there are total of seven prizes)

Applying formula, to find sum of n terms of AP, we get

=

⇒

=

= 700 = 7 (a – 60)

On further specification, we get

 ⇒ 700/7 = a – 60

⇒ 100 + 60 = a

⇒  a = 160

Therefore, value of first prize = Rs 160

Value of second prize = 160 – 20 = Rs 140

Value of third prize = 140 – 20 = Rs 120

Value of fourth prize = 120 – 20 = Rs 100

Value of fifth prize = 100 – 20 = Rs 80

Value of sixth prize = 80 – 20 = Rs 60

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?

Answer:

There are three sections of each class and it is given that the number of trees planted by any class is equal to class number.

The number of trees planted by class I = number of sections

The number of trees planted by class II = number of sections

The number of trees planted by class III = number of sections

Therefore, we have sequence of the form 3, 6, 9 … 12 terms

To find total number of trees planted by all the students, we need to find sum of the sequence 3, 6, 9, 12 … 12 terms.

First term = a = 3, Common difference = d= 6 – 3 = 3 and n = 12

Applying formula, to find sum of n terms of AP , we get

=

= 6 (2+11)

= 6 x 13

= 78

Therefore, number of trees planted by 1 section of classes = 78

number of trees planted by 3 section of classes = 3 x 78 = 234

Therefore, 234 trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centers alternatively at A and B, starting with center 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.

Answer:

Length of Semi-perimeter of circle = πr

Length of semi-circle of radii 0.5 cm = π(0.5) cm = π/2

Length of semi-circle of radii 1.0 cm = π(1.0) cm

Length of semi-circle of radii 1.5 cm = π(1.5) cm = 3π/2

Therefore, we have sequence of the form:

π(0.5), π(1.0), π(1.5) … 13 terms {There are total of thirteen semi–circles}.

To find total length of the spiral, we need to find sum of the sequence π(0.5), π(1.0), π(1.5) … 13 terms

First term = a = 0.5, Common difference = 1.0 – 0.5 = 0.5 and n = 13

Applying formula, to find sum of n terms of AP, we get

=

=

= (13/2) (7π)

= (91π / 2)

= ((91 x 22) / 2) x (7)

= (13 x 11)

= 143

Therefore, the lenght of such spiral of 13 consecutive semi – circles will be 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?

Answer:

The number of logs in the bottom row = 20

The number of logs in the next row = 19

The number of logs in the next to next row = 18

Therefore, we have sequence of the form 20, 19, 18 …

First term = a = 20, Common difference = d = 19 – 20 = – 1

We need to find that how many rows make total of 200 logs.

Applying formula, to find sum of n terms of AP, we get

=

⇒ 400 = n (40 – n + 1)

⇒ 400 = n (41 – n)

⇒ 400 = 41n – n²

⇒ n² – 41n + 400 = 0

⇒ n² – 16n -25n + 400 = 0

⇒ n (n – 16) – 25 (n – 16) = 0

⇒ (n – 16) (n – 25) = 0

Either (n – 16) = 0 or n – 25 = 0

⇒ n = 16 or n = 25

⇒ a_n  = a + (n – 1)d

⇒ a_{16 = 20 + (16 – 1) (- 1)}

⇒ a_{16 = 20 – 15}

⇒ a_{16 = 5}

Similarly,

⇒ a_{25 = 20 + (25 – 1) (- 1)}

⇒ a_{25 = 20 – 24}

 _{= – 4}

_{Clearly, the number of logs in 16th row is 5. However, the number of logs in 25th row is negative, which is not possible.}

Therefore,  200 logs can be placed in 16 rows and the number of logs in 16th row is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 meters from the first potato, and the other potatoes are placed 3 meters 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?

Answer:

The distance of potatoes are as follows :

5, 8, 11, 14, …..

It can be observed that these distances are in A.P.

  a = 5

 d =  8 – 5 = 3

From the formula, we get

= 

= 5 [10 + 9 x 3]

= 5 (10 + 27)

= 5 x 37

= 185

 As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, the total distance that the competitor will be run = 2 x 185 = 370m.

EXERCISE 5.4

1. Which term of the AP: 121, 117, 113, ….. is its first negative term?

Answer:

Given: 121, 117, 113, …….

Here, a = 121 and d = 117-121 = – 4

Let the nth term of the given A.P. be the first negative term. Then, a_n  < 0.

⇒ 121+ (n – 1) x (- 4) < 0                                    [  a_n = a + (n – 1) d]

⇒ 125 – 4n < 0

⇒ – 4n < -125

⇒

Therefore, n = 32

Hence, the first negative term is 32^nd term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of sixteen terms of the AP.

Answer:

We know that,

⇒ a_n = a + (n – 1) d

⇒ a₃ = a + (3 – 1) d

⇒ a₃ = a + 2d

Similarly,   a₇ = a + 6d

Given,      a₃ + a₇ = 6

⇒ (a + 2d) + (a + 6d) = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3

⇒ a = 3 – 4d  (i)

Also, it is given that, (a₃) x (a_{7) = 8}

⇒ (a + 2d)  x (a + 6d) = 8

From Equation (i),

⇒  (3 – 4d + 2d ) x (3 – 4d + 6d) = 8

⇒  (3 – 2d) x (3 + 2d) = 8

⇒   9 – 4d² = 8

⇒   4d² = 9 – 8 = 1

⇒ d² = 1/4

⇒ d = ± 1/2

⇒ d = 1/2 or 1/2

From equation (i)

(When d = 1/2)

⇒ a = 3 – 4d

⇒ a = 3 – 4(1/2)

⇒ 3 – 2 = 1

(When d is – 1/2)

⇒ a = 3 – 4(-1/2)

⇒ a = 3 + 2 = 5

(When a is 1 & d is 1/2)

 

⇒   8 = [2 + 15/2]

⇒  4 (19) = 76

(When a is 5 & d is  – 1/2)

⇒ 8 [10 + (15(-1/2))]

⇒ 8 (5/2)

⇒ 20

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 5/2 m apart, what is the length of the wood required for the rungs?

Answer:

It is given that the rungs are 25 cm apart and the top and bottom rungs are 2 1/2 m apart.

Therefore,Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First term, a = 45

Last term, l = 25

n = 11

⇒ S_n = n/2(a+1)

Therefore,S_{10 = 11/2(45+25) = 11/2 + 70 = 385cm.}

_{Therefore, the length of the wood required for the rungs is 385 cm}.

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.

Answer:

Let there be a value of x  such that the sum of the numbers of the houses preceding the houses numbered x is equal to the sum of the numbers of the houses following it.

That is , 1 + 2 + 3….. + (x + 1) = (x + 1) + (x + 2)………  + 49

Therefore, 1 + 2 + 3….. + (x + 1) = [1 + 2 + ….. + x + (x + 1)…… + 49 ] – [1 + 2 + 3….. + (x)]

Therefore,

⇒  x(x – 1) = 49 x 50 – x(x + 1)

⇒  x(x – 1) + x(x + 1) = 49 x 50

⇒ x² – x + x + x² = 49 x 50

⇒  x² = 49 x 25

Therefore, x = 7 x 5 = 35.

Since, x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of 1/4 m and a tread of 1/2 m (see figure). Calculate the total volume of concrete required to build the terrace.

Answer:

Volume of concrete required to build the first step, second step, third step, ……. (in m2) are

From the figure, it can be observed that

1st step is 1/2 m wide,

2nd step is 1 m wide,

3rd step is 3/2 m wide,

Therefore, the width of each step is increasing by 1/2 m each time whereas their height 1/4 m and length

50 m remains the same.

Therefore, the widths of these steps are

1/2,1, 3/2, 2,…

Volume of concrete in 1st step =  

Volume of concrete in 2nd step = 

Volume of concrete in 3rd step = 

It can be observed that the volumes of concrete in these steps are in an A.P.

 

⇒ a = 25/4

⇒ d= 25/2 – 25/4 = 25/4

and  ,

=

=

= 15/2(100) = 750

Volume of concrete required to build the terrace is 750 m³.