Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2

Question 1.
Write the first 6 terms of the sequence whose nth terms are given below and classify them as Arithmetic progression Geometric progression, Arithmetic – geometric progression, Harmonic progression and none of them.
(i) \(\frac{1}{2^{n+1}}\)
Answer:

(ii)
Answer:

(iii) \(4\left(\frac{1}{2}\right)^{n}\)
Answer:

(iv) \(\frac{(-1)^{n}}{n}\)
Answer:

(v) \(\frac{2 n+3}{3 n+4}\)
Answer:

(vi) 2018
Answer:
The n^th term a_n = 2018
a₁ = 2018,
a₂ = 2018,
a₃ = 2018,
a₄ = 2018,
a₅ = 2018,
a₆ = 2018,
∴ The given sequence is 2018, 2018, 2018, 2018, 2018, 2018, ………….
This is a œnstant sequence which has same common ratio and common difference.
Hence this is an A. P, G . P and AGP.

(vii) \(\frac{3 n-2}{3^{n-1}}\)
Answer:

Question 2.
Write the first 6 terms of the sequences whose n^th term a_n is given below
(i)
n = 1, a_n = n + 1, a₁ = 1 + 1 = 2
n = 2, a_n = n, a₂ = 2
n = 3, a_n = n + 1, a₃ = 3 + 1 = 4
n = 4, a_n = n, a₄ = 4
n = 5, a_n = n + 1, a₅ = 5 + 1 = 6
n = 6, a_n = n, a₆ = 6
∴ The first six terms are 2, 2, 4, 4, 6, 6

(ii)
n = 1, a₁ = 1, n = 2, a₂ = 2
n = 3, a_n = a_n+1 + a_n-2, a₃ = a_{3 – 1} + a_{3 – 2} = a₂ + a₁ = 2 + 1 = 3

n = 4, a_n = a_n+1 + a_n-2, a₄ = a_{4 – 1} + a_{4 – 2} = a₃ + a₂ = 3 + 2 = 5

n = 5, a_n = a_n+1 + a_n-2, a₅ = a_{5 – 1} + a_{5 – 2} = a₄ + a₃ = 5 + 3 = 8

n = 6, a_n = a_n+1 + a_n-2, a₆ = a_{6 – 1} + a_{6 – 2} = a₅ + a₄ = 8 + 5 = 13
∴ The first six terms are 1, 2, 3, 5, 8, 13

(iii)
Answer:
n = 1, a_n = n, a₁ = 1
n = 2, a_n = n, a₂ = 1
n = 3, a_n = n, a₃ = 1

n = 4, a_n = a_n-1 + a_n-2 + a_n-3
a₄ = a_4-1 + a_4-2 + a_4-3
a₄ = a₃ + a₂ + a₁
a₄ = 3 + 2 + 1 = 6

n = 5, a_n = a_n-1 + a_n-2 + a_n-3
a₅ = a_5-1 + a_5-2 + a_5-3
a₅ = a₄ + a₃ + a₂
a₅ = 6 + 3 + 2 = 11

n = 6, a_n = a_n-1 + a_n-2 + a_n-3
a₆ = a_6-1 + a_6-2 + a_6-3
a₆ = a₅ + a₄ + a₃
a₆ = 11 + 6 + 3 = 20
∴ The first six terms are 1, 2, 3, 6, 11, 20

Question 3.
Write the n^th term of the following sequences.
(i) 2, 2, 4, 4, 6, 6, ……………..
Answer:
The odd terms are a₁ = 2, a₃ = 4, a₅ = 6
The even terms are a₂ = 2, a₄ = 4, a₆ = 6

(ii)
Answer:

The terms in the numerator are 1 , 2 , 3, 4
a = 1 , d = 2 – 1 = 1
a_n = a + (n – 1) d
a_n = 1 + (n – 1)(1) = 1 + n – 1 = n
a_n = n
The terms in the denominator are 2 , 3 , 4 , 5 , 6 .
a = 2, d = 3 – 2 = 1
a_n = a + (n – 1) d
a_n = 2 + (n – 1) (1) = 2 + n – 1 = n + 1
a_n = n + 1
∴ The n^th term of the given sequence is a_n = \(\frac{n}{n+1}\) for all n ∈ N

(iii)
Answer:

The terms in the numerator are 1, 3, 5, 7, 9, ………….
a = 1 , d = 3 – 1 = 2
a_n = a + (n – 1) d
a_n = 1 + (n – 1)2
a_n = 1 + 2n – 2 = 2n – 1
The terms in the denominator are 2, 4, 6, 8, 10, …………..
a = 2, d = 4 – 2 = 2
a_n = a + (n – 1) d
a_n = 2 + (n – 1)(2)
a_n = 2 + 2n – 2 = 2n
∴ The n^th term of the given sequence is a_n = \(\frac{2 n-1}{2 n}\) for all n ∈ N

(iv) 6, 10, 4, 12, 2, 14, 0, 16, – 2 …………………
Answer:
The given sequence is 6, 10, 4, 12, 2, 14, 0, 16, – 2 ……………..
The odd terms are a₁ = 6, a₃ = 4 , a₅ = 2 , a₇ = 0, a₉ = – 2
∴ a_n = n – 7, n is odd
The even terms are a₂ = 10, a₄ = 12 , a₆ = 14 , a₈ = 16
∴ a_n = 8 + n, n is even.

Question 4.
The product of three increasing numbers in a G.P is 5832 . If we add 6 to the second number and 9 to the third number, then resulting numbers form an A.P. Find the numbers in G.P.
Answer:
Let the increasing numbers in G.P be \(\), a, ar.
Given \(\frac{a}{r}\) × a × ar = 5832 ⇒ a³ = 5832 = 18³ ⇒ a = 18
Also given \(\frac{a}{r}\), a + 6, ar + 9 form an A.P.
∴ 2(a + b) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) + (a + 6) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) – \(\frac{a}{r}\) = (ar + 9) – (a + 6)
⇒ a + 6 – \(\frac{a}{r}\) = ar + 9 – a – 6
⇒ a + 6 – \(\frac{a}{r}\) = ar – a + 3
Substituting the value of a = 18, we get


39r = 18r² + 18
18r² – 39r + 18 = 0
(2r – 3)(3r -2) = 0
2r – 3 = 0 or 3r – 2 = 0
r = \(\frac{3}{2}\) or r = \(\frac{2}{3}\)

Case (i) When a = 18, r = \(\frac{3}{2}\) the numbers in G.P are

Case (ii) When a = 18, r = \(\frac{2}{3}\)

Question 5.
Write the n^th term of the sequence \(\frac{3}{1^{2} \cdot 2^{2}}, \frac{5}{2^{2} \cdot 3^{2}}, \frac{7}{3^{2} \cdot 4^{2}}\), …………….. as a difference of two terms.
Answer:

The terms in the numerator are 3,5, 7
which forms an A. P with first term a = 3 and common difference d = 5 – 3 = 2
n^th term t_n = a + (n – 1) d
= 3 + (n – 1)(2)
= 3 + 2n – 2 = 2n + 1
t_n = 2n + 1
The terms in the denominator are 1² . 2², 2² . 3², 3² . 4² ……………….
n^th term t_n = n² . (n + 1)²

Question 6.
If t_k is the k^th term of a G.P then show that t_{n – k}, t_k, t_{n + k} also form a G.F for any positive integer k.
Answer:
Given t^k is the k^th term of a G.P. We have n^th term of a G.P is t_n = ar^n-1

Question 7.
If a, b, c are in geometric progression and if a^1/x = b^1/y = c^1/z are in Arithmetic progression.
Answer:

Question 8.
The A.M of two numbers exceeds their G.M by 10 and H.M by 16. Find the numbers.
Answer:
Let the numbers be a and b

Question 9.
If the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal then show that p , q and r are in A. P.
Answer:
The roots are equal ⇒ ∆ = 0
(i.e.) b² – 4ac = 0
Hence, a = q – r ; b = r – p ; c = p – q
b² – 4ac = 0
⇒ (r – p)² – 4(q – r)(p – q) = 0
r² + p² – 2pr – 4[qr – q² – pr + pq] = 0
r² + p² – 2pr – 4qr + 4q² + 4pr – 4pq = 0
(i.e.) p² + 4q² + r² – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r)² = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.

Question 10.
If a , b , c are respectively the p^th, q^th and r^th terms of a G . P show that (q – r) log a + (r – p) log b + (p – q) log c = 0
Answer:
Let A be first term and R be the jmnon ratio of the G.P.
Given a = p^th term of the G.P
General term of a G. P with first term A and common ratio R is t_n = AR^{n – 1}
∴ a = t_p = AR^{P – 1}
log a = log AR^p-1 = log A + log R^p-1 = log A + (p – 1) log R

b = q^th term of the G.P
b = t_q = AR^q-1
log b = log AR^q-1 = log A + log R^q-1 = log A + (q – r)log R

c = r^th term of the G.P
c = t_r = AR^r-1
log c = log AR^r-1 = log A + log R^r-1 = log A + (r – 1) log R

(q – r) log a + (r – p) log b + (p – q) log c
= (q – r) [log A + (p – 1) log R] + (r – p) [ log A + (q – 1) log R ] + (P – q) [ log A + (r – 1) log R]
= (q – r) log A + (q – r) (p – 1 ) log R + (r – p) log A + (r – p) (q – 1) log R + (P – q) log A + (p – q) (r – 1 ) log R
= [ q – r + r – p + p – q ] log A + [ (q – r) (p – 1) + (r – p) (q – 1) + (p – q)(r – 1)] log R
= 0 × log A + [pq – q – rp + r + rq – r – pq + p + pr – p – rq + q] log R
= 0 × log R = 0
∴ (q – r) log a + (r – p) log b + (p – q) log c = 0