Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.5

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.5

Question 1.
Check whether the following sequences are in A.P.?

(i) a – 3, a – 5, a – 7,…
Answer:
a – 3, a – 5, a – 7…….
t₂ – t₁ = a – 5 – (a – 3)
= a – 5 – a + 3
= -2
t₃ – t₂ = a – 7 – (a – 5)
= a – 7 – a + 5
= -2
t₂ – t₁ = t₃ – t₂
(common difference is same)
The sequence is in A.P.

(ii) \(\frac { 1 }{ 2 } \), \(\frac { 1 }{ 3 } \), \(\frac { 1 }{ 4 } \), \(\frac { 1 }{ 5 } \), ……….
Answer:
t₂ – t₁ = \(\frac { 1 }{ 3 } \) – \(\frac { 1 }{ 2 } \) = \(\frac { 2-3 }{ 6 } \) = \(\frac { -1 }{ 6 } \)
t₃ – t₂ = \(\frac { 1 }{ 4 } \) – \(\frac { 1 }{ 3 } \) = \(\frac { 3-4 }{ 12 } \) = \(\frac { -1 }{ 12 } \)
t₂ – t₁ ≠ t₃ – t₂
The sequence is not in A.P.

(iii) 9, 13, 17, 21, 25,…
Answer:
t₂ – t₁ = 13 – 9 = 4
t₃ – t₂ = 17 – 13 = 4
t₄ – t₃ = 21 – 17 = 4
t₅ – t₄ = 25 – 21 = 4
Common difference are equal
∴ The sequence is in A.P.

(iv) \(\frac { -1 }{ 3 } \), 0, \(\frac { 1 }{ 3 } \), \(\frac { 2 }{ 3 } \)
t₂ – t₁ = 0 – (-\(\frac { 1 }{ 3 } \))
= 0 + \(\frac { 1 }{ 3 } \) = \(\frac { 1 }{ 3 } \)
t₃ – t₂ = \(\frac { 1 }{ 3 } \) – 0 = \(\frac { 1 }{ 3 } \)
t₂ – t₁ = t₃ – t₂
The sequence is in A.P.

(v) 1,-1, 1,-1, 1, -1, …
t₂ – t₁ = -1 – 1 = -2
t₃ – t₂ = 1 – (-1) = 1 + 1 = 2
t₄ – t₃ = -1-(1) = – 1 – 1 = – 2
t₅ – t₄ = 1 – (-1) = 1 + 1 = 2
Common difference are not equal
∴ The sequence is not an A.P.

Question 2.
First term a and common difference d are given below. Find the corresponding A.P. ?
(i) a = 5 ,d = 6
Answer:
Here a = 5,d = 6
The general form of the A.P is a, a + d, a + 2d, a + 3d….
The A.P. 5, 11, 17, 23 ….

(ii) a = 7, d = -5
Answer:
The general form of the A.P is a, a + d,
a + 2d, a + 3d… .
The A.P. 7, 2, -3, -8 ….

(iii) a = \(\frac { 3 }{ 4 } \), d = \(\frac { 1 }{ 2 } \)
Answer:
The general form of the A.P is a, a + d, a + 2d, a + 3d….
\(\frac { 3 }{ 4 } \),\(\frac { 3 }{ 4 } \) + \(\frac { 1 }{ 2 } \),\(\frac { 3 }{ 4 } \) + 2(\(\frac { 1 }{ 2 } \)), \(\frac { 3 }{ 4 } \) + 3 (\(\frac { 1 }{ 2 } \))
The A.P. \(\frac { 3 }{ 4 } \), \(\frac { 5 }{ 4 } \), \(\frac { 7 }{ 4 } \), …….

Question 3.
Find the first term and common difference of the Arithmetic Progressions whose n^th terms are given below
(i) t_n = -3 + 2n
(ii) t_n = 4 – 7n
Solution:
(i) a = t₁ = -3 + 2(1) = -3 + 2 = -1
d = t₂ – t₁
Here t₂ = -3 + 2(2) = -3 + 4 = 1
∴ d = t₂ – t₁ = 1 – (-1) = 2
(ii) a = t₁ = 4 – 7(1) = 4 – 7 = -3
d = t₂ – t₁
Here t₂ = 4 – 7(2) = 4 – 14 – 10
∴ d = t₂ – t₁ = 10 – (-3) = -7

Question 4.
Find the 19^th term of an A.P. -11, -15, -19,…
Answer:
First term (a) = -11
Common difference (d) = -15 -(-11)
= -15 + 11 = -4
n = 19
t_n = a + (n – 1) d
t_n = -11 + 18(-4)
= -11 – 72
t₁₉ = -83
19^th term of an A.P. is – 83

Question 5.
Which term of an A.P. 16, 11, 6,1, ……….. is -54?
Solution:
A.P = 16, 11,6, 1, ………..
It is given that
t_n = -54
a = 16, d = t₂ – t₁ = 11 – 16 = -5
∴ t_n = a + (n – 1)d
-54 = 16 + (n – 1) (-5)
-54 = 16 – 5n + 5
21 – 5n = -54
-5n = -54 -21
-5n = -75
n = \(\frac { 75 }{ 5 } \) =15
∴ 15th term is -54.

Question 6.
Find the middle term(s) of an A.P. 9, 15, 21, 27, …, 183.
Answer:
First term (a) = 9
Last term (l) = 183
Common difference (d) = 15 – 9 = 6
n = \(\frac { l-a }{ d } \) + 1
= \(\frac { 183-9 }{ 6 } \) + 1
= \(\frac { 174 }{ 6 } \) + 1
= 29 + 1
= 30
middle term = 15^th term of
16^th term
t_n = a + (n – 1)d
t₁₅ = 9 + 14(6)
= 9 + 84 = 93
t₁₆ = 9 + 15(6)
= 9 + 90 = 99
The middle term is 93 or 99

Question 7.
If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.
Solution:
Nine times ninth term = Fifteen times fifteenth term
9t₉ = 15t₁₅
9(a + 8d) = 5(a + 14d)
9a + 72d = 15a + 210
15a + 210d – 9a – 72d = 0
⇒ 6a + 138 d = 0
⇒ 6(a + 23 d) = 0
⇒ 6(a + (24 – 1)d) = 0
⇒ 6t₂₄ = 0. Hence it is proved.

Question 8.
If 3 + k, 18 – k, 5k + 1 are in A.P. then find k?
Answer:
3 + k, 18 – k, 5k + 1 are in AP
∴ t₂ – t₁ = t₃ – t₂ (common difference is same)
18 – k – (3 + k) = 5k + 1 – (18 – k)
18 – k – 3 – k = 5k + 1 – 18 + k
15 – 2k = 6k – 17
32 = 8k
k = \(\frac { 32 }{ 8 } \) = 4
The value of k = 4

Question 9.
Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
Solution:
A.P = x, 10, y, 24, z,…
d = t₂ – t₁ = 10 – x ………….. (1)
= t₃ – t₂ = y – 10 ………….. (2)
= t₄ – t₃ = 24 – y …………. (3)
= t₅ – t₄ = z – 24 ………….. (4)
(2) and (3)
⇒ y – 10 = 24 – y
2y = 24 + 10 = 34
y = \(\frac { 34 }{ 2 } \) = 17
(1) and (2)
⇒ 10 – x = y – 10
10 – x = 17 – 10 = 7
-x = 7 – 10
-x = -3 ⇒ x = 3
From (3) and (4)
24 – y = z – 24
24 – 17 = z – 24
7 = z – 24
∴ z = 7 + 24 = 31
∴ Solutions x = 3
y = 17
z = 31

Question 10.
In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?
Answer:
Number of seats in the first row
(a) = 20
∴ t₁ = 20
Number of seats in the second row
(t₂) = 20 + 2
= 22
Number of seats in the third row
(t₃) = 22 + 2
= 24
Here a = 20 ; d = 2
Number of rows
(n) = 30
t_n = a + (n – 1)d
t₃₀ = 20 + 29(2)
= 20 + 58
t₃₀ = 78
Number of seats in the last row is 78

Question 11.
The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Solution:
Let the three consecutive terms be a – d, a, a + d
Their sum = a – d + a + a + d = 27
3a = 27
a = \(\frac{27}{3}\) = 9
Their product = (a – d)(a)(a + d) = 288
= 9(a² – d²) = 288
⇒ 9(9 – d²) = 288
⇒ 9(81 – d²) = 288
81 – d² = 32
-d² = 32 – 81
d² = 49
⇒ d = ± 7
∴ The three terms are if a = 9, d = 7
a – d, a , a + d = 9 – 7, 9 + 7
A.P. = 2, 9, 16
if a = 9, d = -7
A.P. = 9 – (-7), 9, 9 + (-7)
= 16, 9, 2

Question 12.
The ratio of 6^th and 8^th term of an A.P is 7:9. Find the ratio of 9^th term to 13^th term.
Answer:
Given : t₆ : t₈ = 7 : 9 (using t_n = a + (n – 1)d
a + 5d : a + 7d = 7 : 9
9 (a + 5 d) = 7 (a + 7d)
9a + 45 d = 7a + 49d
9a – 7a = 49d – 45d
2a = 4d
a = 2d
To find t₉ : t₁₃
t₉ : t₁₃ = a + 8d : a + 12d
= 2d + 8d : 2d + 12d
= 10d : 14d
= 5 : 7
∴ t₉ : t₁₃ = 5 : 7

Question 13.
In a winter season, the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days.
Answer:
Let the five days temperature be
(a – 2d), (a – d), a,(a + d) and (a + 2d)
Sum of first three days temperature = 0
a – 2d + a – d + a = 0
3a – 3d = 0
a – d = 0 …..(1)
Sum of the last three days temperature = 18°C
a + a + d + a + 2d = 18
3a + 3d = 18
(÷ by 3) ⇒ a + d = 6 ……(2)
By adding (1) and (2)

Substitute to value of a = 3 in (2)
d = 3
The temperature in 5 days are
(3 – 6), (3 – 3), 3, (3 + 3) and (3 + 6)
-3°C, 0°C, 3°C, 6°C, 9°C

Question 14.
Priya earned ₹15,000 in the first month. Thereafter her salary increases by ₹1500 per year. Her expenses are ₹13,000 during the first year and the expenses increases by ₹900 per year. How long will it take her to save ₹20,000 per month.
Answer:
Tabulate the given table

Monthly savings form an A.P.
2000, 2600, 3200 …..
a = 2000; d = 2600 – 2000 = 600
Given t_n = 20,000
t_n = a + (n – 1) d
20000 = 2000 + (n – 1) 600
20000 = 2000 + 600n – 600
= 1400 + 600n
20000 – 1400 = 600n
18600 = 600n
n = \(\frac { 18600 }{ 600 } \) = 31
He will take 31 years to save ₹ 20,000 per month