Samacheer Kalvi 10th Maths Book Back Solution:
Tamil Nadu 10th Maths Book Back Answers Unit 2 – Numbers and Sequences Ex 2.5 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.
The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 2 Numbers and Sequences consists of 5 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Numbers and Sequences Ex 2.5 Answers in English.
10th Maths Book Back Answers/Solutions:
TN Samacheer Kalvi 10th Maths Unit 2 Chapter 5 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,
Chapter 2
Numbers and Sequences Ex 2.5
1.Check whether the following sequences are in A.P.
(i) a – 3, a – 5, a – 7, ………
(ii) 12,13,14,14, ………..
(iii) 9, 13, 17, 21, 25, ………
(iv) −13,0,13,23,………..
(v) 1,-1, 1,-1, 1,-1,…
Solution:
To prove it is an A.P, we have to show d = t2 – t1 = t3 – t2.
(i) a – 3, a – 5, a – 7………
t1,t2,t3
d = t2 – t1 = a – 5 – (a – 3) = a – 5 – a + 3 = -2
∴ d = -2 ∴ It is an A.P.
d = t3 – t2 = a – 7 – (a – 5) = a – 7 – a + 5 = -2
(v) 1,-1, 1,-1, 1,-1,…
d = t2 – t1 = -1 -1 = -2
d = t3 – t2 = 1 – (-1) = 2
-2 ≠ 2 ∴ It is not an A.P.
2.First term a and common difference d are given below. Find the corresponding A.P.
(i) a = 5, d = 6
(ii) a = 7, d = 5
(iii) a = 34, d = 12
Solution:
(i) a = 5, d = 6
A.P a, a + d, a + 2d, ………
= 5, 5 + 6, 5 + 2 × 6, ………
= 5, 11, 17,…
(ii) a = 7,d = -5
A.P. = a,a + d,a + 2d,…
= 7,7 + (-5), 7 + 2(-5), ……….
= 7, 2, -3, …….,…
(iii) a = 34, d = 12
3.Find the first term and common difference of the Arithmetic Progressions whose nthterms are given below
(i) tn = -3 + 2n
Answer:
tn = -3 + 2 n
t1 = -3 + 2(1) = -3 + 2
= -1
t2 = -3 + 2(2) = -3 + 4
= 1
First term (a) = -1 and
Common difference
(d) = 1 – (-1) = 1 + 1 = 2
(ii) tn = 4 – 7n
Answer:
tn = 4 – 7n
t1 = 4 – 7(1)
= 4 – 7 = -3
t2 = 4 – 7(2)
= 4 – 14 = -10
First term (a) = – 3 and
Common difference (d) = 10 – (-3)
= – 10 + 3
= – 7
4.Find the 19th term of an A.P. -11, -15, -19, ………..
Solution:
A.P = -11, -15, -19, ……..
a = -11
d = t2 – t1 =-15-(-11)
= -15 + 11
= -4
n = 19
∴ tn = a + (n – 1)d
t19 = -11 + (19 – 1)(-4)
= -11 + 18 × -4
= -11 – 72
= -83
5.Which term of an A.P. 16, 11, 6, 1,… is -54?
Answer:
First term (a) = 16
Common difference (d) = 11 – 16 = -5
tn = – 54
a + (n – 1) d = -54
16 + (n – 1) (-5) = -54
54 + 21 = -54
54 + 21 = 5n
75 = 5n
n = 755 = 15
The 15th term is – 54
6.Find the middle term(s) of an A.P. 9, 15, 21, 27, ……. ,183.
Solution:
A.P = 9, 15, 21, 27,…, 183
No. of terms in an A.P. is
n = l−ad + 1
a = 9, l = 183, d = 15 – 9 = 6
∴ n = 183−96 + 1
= 1746 + 1
= 29 + 1 = 30
∴ No. of terms = 30. The middle must be 15th term and 16th term.
∴ t15 = a + (n – 1)d
= 9 + 14 × 6
=9 + 84
= 93
t16 = a + 15 d
= 9 + 15 × 6
= 9 + 90 = 99
∴ The middle terms are 93, 99.
7.If nine times the ninth term is equal to the fifteen times fifteenth term, Show that six times twenty fourth term is zero.
Answer:
tn = a + (n – 1)d
9 times 9th term = 15 times 15th term
9t9 = 15 t15
9[a + 8d] = 15[a + 14d]
9a + 72d = 15a + 210d
9a – 15a + 72 d – 210 d = 0
-6a – 138 d = 0
6a + 138 d = 0
6 [a + 23 d] = 0
6 [a + (24 – 1)d] = 0
6 t24 = 0
∴ Six times 24th terms is 0.
8.If 3 + k, 18 – k, 5k + 1 are in A.P. then find k.
Solution:
3 + k, 18 – k, 5k + 1 are in A.P
⇒ 2b = a + c if a, b, c are in A.P
9.Find x,y and z gave that the numbers x,
10, y, 24, z are in A.P.
Answer:
x, 10, y, 24, z are in A.P
t2 – t1 = 10 – x
d = 10 – x …..(1)
t3 – t2 = y – 10
d = y – 10 ……(2)
t4 – t3 = 24 – y
d = 24 – y …..(3)
t5 – t4 = z – 24
d = z – 24 …..(4)
From (2) and (3) we get
y – 10 = 24 – y
2y = 24 + 10
2y = 34
y = 17
From (1) and (2) we get
10 – x = y – 10
– x – y = -10 -10
-x -y = -20
x + y = 20
x + 17 = 20(y = 17)
x = 20 – 17 = 3
From (1) and (4) we get
z – 24 = 10 – x
z – 24 = 10 – 3 (x = 3)
z – 24 = 7
z = 7 + 24
z = 31
The value of x = 3, y = 17 and z = 31
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?
Solution:
t1 = a = 20
t2 = a + 2 = 22
t3 = a + 2 + 2 = 24 ⇒ d = 2
∴ There are 30 rows.
t30 = a + 29d
= 20 + 29 × 2
= 20 + 58
= 78
∴ There will be 78 seats in the last row.
11.The Sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Answer:
Let the three consecutive terms be a – d, a and a + d
By the given first condition
a – d + a + a + d = 27
3a = 27
a = 273 = 9
Again by the second condition
(a – d) (a) (a + d) = 288
a (a2 – d2) = 288
9(81 – d2) = 288 (a = 9)
81 – d2 = 2889
81 – d2 = 32
∴ d2 = 81 – 32
= 49
d = 49−−√ = ± 7
When a = 9, d = 7
a + d = 9 + 7 = 16
a = 9
a – d = 9 – 7 = 2
When a = 9, d = -7
a + d = 9 – 7 = 2
a = 9
a – d = 9 – (-7) = 9 + 7 = 16
The three terms are 2, 9, 16 (or) 16, 9, 2
12.The ratio of 6th and 8th term of an A.P is 7:9. Find the ratio of 9th term to 13th term.
Solution:
t6t8=79
a+5da+7d = 79
9a + 45d = 7a + 49d
9 a + 45 – 7d = 7a + 49 d
9a + 45d – 7a – 49d = 0
2a – 4d = 0 ⇒ 2a = 4d
a = 2d
Substitue a = 2d in
13.In a winter season let us take 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.
Solution:
Let the five days temperature be (a – d), a, a + d, a + 2d, a + 3d.
The three days sum = a – d + a + a + d = 0
⇒ 3a = 0 ⇒ a = 0. (given)
a + d + a + 2d + a + 3d = 18
3a + 6d = 18
3(0) + 6 d = 18
6d = 18
d = 186 = 3
∴ The temperature of each five days is a – d, a, a + d, a + 2d, a + 3d
0 – 3, 0, 0 + 3, 0 + 2(3), 0 + 3(3) = -3°C, 0°C, 3°C, 6°C, 9°C
14.Priya earned ₹ 15,000 in the first month. Thereafter her salary increased 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 for her to save ₹20,000 per month.
Solution:
We find that the yearly savings is in A.P with a1 = 2000 and d = 600.
We are required to find how many years are required to save 20,000 a year …………..
an = 20,000
an = a + (n – 1)d
20000 = 2000 + (n – 1)600
(n – 1)600 = 18000
n – 1 = 18000600 = 30
n = 31 years
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