Samacheer Kalvi 10th Maths Book Back Solution:
Tamil Nadu 10th Maths Book Back Answers Unit 2 – Numbers and Sequences Ex 2.4 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 2.4 Answers in English.
10th Maths Book Back Answers/Solutions:
TN Samacheer Kalvi 10th Maths Unit 2 Chapter 4 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,
Chapter 2
Numbers and Sequences Ex 2.4
1.Find the next three terms of the following sequence.
(i) 8, 24, 72, …….
(ii) 5, 1, -3, …….
(iii) 14,29,316………..
Solution:
(i) 8, 24, 72…
In an arithmetic sequence a = 8,
d = t1 – t1 = t3 – t2
= 24 – 8 72 – 24
= 16 ≠ 48
So, it is not an arithmetic sequence. In a geometric sequence,
r = t2t1=t3t2
⇒ 248=7224
⇒ 3 = 3
∴ It is a geometric sequence
∴ The nth term of a G.P is tn = arn-1
∴ t4 = 8 × 34-1
= 8 × 33
= 8 × 27
= 216
t5 = 8 × 35-1
= 8 × 34
= 8 × 81
= 648
t6 = 8 × 36-1
= 8 × 35
= 8 × 243
= 1944
The next 3 terms are 8, 24, 72, 216, 648, 1944.
(ii) 5, 1, -3, …
d = t2 – t1 = t3 – t2
⇒ 1 – 5 = -3-1
-4 = -4 ∴ It is an A.P.
tn = a+(n – 1)d
t4 = 5 + 3 × – 4
= 5 – 12
= -7
15 = a + 4d
= 5 + 4 × -4
= 5 – 16
= -11
t6 = a + 5d
= 5 + 5 × – 4
= 5 – 20
= – 15
∴ The next three terms are 5, 1, -3, -7, -11, -15.
(iii) 14,29,316,………..
Here an = Numerators are natural numbers and denominators are squares of the next numbers
14,29,316,425,536,649………….
2.Find the first four terms of the sequences whose nth terms are given by
(i) an = n3 -2
Answer:
an = n3 – 2
a1 = 13 – 2 = 1 – 2 = -1
a2 = 23 – 2 = 8 – 2 = 6
a3 = 33 – 2 = 27 – 2 = 25
a4 = 43 – 2 = 64 – 2 = 62
The four terms are -1, 6, 25 and 62
(ii) an = (-1)n+1 n(n + 1)
Answer:
an = (-1)n+1 n(n + 1)
a1 = (-1)2 (1) (2) = 1 × 1 × 2 = 2
a2 = (-1)3 (2) (3) = -1 × 2 × 3 = -6
a3 = (-1)4 (3) (4) = 1 × 3 × 4 = 12
a4 = (-1)5 (4) (5) = -1 × 4 × 5 = -20
The four terms are 2, -6, 12 and -20
(iii) an = 2n2 – 6
Answer:
an = 2 n2 – 6
a1 = 2(1)2 – 6 = 2 – 6 = -4
a2 = 2(2)2 – 6 = 8 – 6 = 2
a3 = 2(3)2 – 6 = 18 – 6 = 12
a4 = 2(4)2 – 6 = 32 – 6 = 26
The four terms are -4, 2, 12, 26
3.Find the nth term of the following sequences
(i) 2, 5, 10, 17, ……….
(ii) 0, 12, 23,…..
(iii) 3, 8, 13, 18, ………
Solution:
(i) 2, 5, 10, 17
= 12 + 1, 22 + 1, 32 + 1, 42 + 1 ……….
∴ nth term is n2+1
(ii) 0, 12,23,………….
= 1−11,2−12,3−13…..
⇒ n−1n
∴ nth term is n−1n
(iii) 3, 8, 13, 18
a = 3
d = 5
tn = a + (n – 1)d
= 3 + (n – 1)5
= 3 + 5n – 5
= 5n – 2
∴ nth term is 5n – 2
4.Find the indicated terms of the sequences whose nth terms are given by
(i) an = 5nn+2 ; a6 and a13
(ii) an = -(n2 – 4); a4 and a11
Solution:
5.Find a8 and a15 whose nth term is
6.If a1 = 1, a2 = 1 and an = 2an-1 + an-2 n > 3, n ∈ N. Then find the first six terms of the sequence.
Answer:
a1 = a2 = 1
an = 2an-1 + an-2
a3 = 2a3-1 + a3-2 = 2a2 + a1
= 2(1) + 1 = 3
a4 = 2a4-1 + a4-2
= 2a3 + a2
= 2(3) + 1 = 6 + 1 = 7
a5 = 2 a5-1 + a5-2
= 2a4 + a3
= 2(7) + 3 = 17
a6 = 2a6-1 + a6-2
= 2a5 + a4
= 2(17) + 7
= 34 + 7 = 41
The sequence is 1, 1, 3, 7, 17,41, …
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