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

a_{1} = 2018,

a_{2} = 2018,

a_{3} = 2018,

a_{4} = 2018,

a_{5} = 2018,

a_{6} = 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, a_{1} = 1 + 1 = 2

n = 2, a_{n} = n, a_{2} = 2

n = 3, a_{n} = n + 1, a_{3} = 3 + 1 = 4

n = 4, a_{n} = n, a_{4} = 4

n = 5, a_{n} = n + 1, a_{5} = 5 + 1 = 6

n = 6, a_{n} = n, a_{6} = 6

∴ The first six terms are 2, 2, 4, 4, 6, 6

(ii)

n = 1, a_{1} = 1, n = 2, a_{2} = 2

n = 3, a_{n} = a_{n+1} + a_{n-2}, a_{3} = a_{3 – 1} + a_{3 – 2} = a_{2} + a_{1} = 2 + 1 = 3

n = 4, a_{n} = a_{n+1} + a_{n-2}, a_{4} = a_{4 – 1} + a_{4 – 2} = a_{3} + a_{2} = 3 + 2 = 5

n = 5, a_{n} = a_{n+1} + a_{n-2}, a_{5} = a_{5 – 1} + a_{5 – 2} = a_{4} + a_{3} = 5 + 3 = 8

n = 6, a_{n} = a_{n+1} + a_{n-2}, a_{6} = a_{6 – 1} + a_{6 – 2} = a_{5} + a_{4} = 8 + 5 = 13

∴ The first six terms are 1, 2, 3, 5, 8, 13

(iii)

Answer:

n = 1, a_{n} = n, a_{1} = 1

n = 2, a_{n} = n, a_{2} = 1

n = 3, a_{n} = n, a_{3} = 1

n = 4, a_{n} = a_{n-1} + a_{n-2} + a_{n-3}

a_{4} = a_{4-1} + a_{4-2} + a_{4-3}

a_{4} = a_{3} + a_{2} + a_{1}

a_{4} = 3 + 2 + 1 = 6

n = 5, a_{n} = a_{n-1} + a_{n-2} + a_{n-3}

a_{5} = a_{5-1} + a_{5-2} + a_{5-3}

a_{5} = a_{4} + a_{3} + a_{2}

a_{5} = 6 + 3 + 2 = 11

n = 6, a_{n} = a_{n-1} + a_{n-2} + a_{n-3}

a_{6} = a_{6-1} + a_{6-2} + a_{6-3}

a_{6} = a_{5} + a_{4} + a_{3}

a_{6} = 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 a_{1} = 2, a_{3} = 4, a_{5} = 6

The even terms are a_{2} = 2, a_{4} = 4, a_{6} = 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 a_{1} = 6, a_{3} = 4 , a_{5} = 2 , a_{7} = 0, a_{9} = – 2

∴ a_{n} = n – 7, n is odd

The even terms are a_{2} = 10, a_{4} = 12 , a_{6} = 14 , a_{8} = 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 ⇒ a^{3} = 5832 = 18^{3} ⇒ 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 = 18r^{2} + 18

18r^{2} – 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 1^{2} . 2^{2}, 2^{2} . 3^{2}, 3^{2} . 4^{2} ……………….

n^{th} term t_{n} = n^{2} . (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)x^{2} + (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.) b^{2} – 4ac = 0

Hence, a = q – r ; b = r – p ; c = p – q

b^{2} – 4ac = 0

⇒ (r – p)^{2} – 4(q – r)(p – q) = 0

r^{2} + p^{2} – 2pr – 4[qr – q^{2} – pr + pq] = 0

r^{2} + p^{2} – 2pr – 4qr + 4q^{2} + 4pr – 4pq = 0

(i.e.) p^{2} + 4q^{2} + r^{2} – 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