Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 10 Sequences and Series Ex 10(a) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 10 Sequences and Series Exercise 10(a)

Question 1.

Which of the following in a sequence?

(i) f(x) = [x], x ∈ R

(ii) f(x) = |x|, x ∈ R

(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N

Solution:

(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N is a sequence f(n) : N → X, X ⊂ R.

Question 2.

Determine if (t_{n}) is an arithmetic sequence if :

(i) t_{n} = an^{2} + bn

Solution:

t_{n} = an^{2} + bn

⇒ t_{n+1} = a(n + 1)^{2} + b(n – 1)

⇒ t_{n+1} – t_{n} = a{(n + 1)^{2} – n^{2}} + b{n + 1 – n}

= a(2n + 1) + b

Which is not independent of n.

∴ (t_{n}) is not an A.P.

(ii) t_{n} = an + b

Solution:

t_{n} = an + b

⇒ t_{n+1} = a(n + 1) + b

Now t_{n+1} – t_{n}

= {a(n + 1) + b} – {an + b}

= a (constant)

∴ (t_{n}) is an arithmetic sequence.

(iii) t_{n} = an^{2} + b

Solution:

t_{n} = an^{2} + b

⇒ t_{n+1} = a(n + 1)^{2} + b

∴ t_{n+1} – t_{n} = a[(n + 1)^{2} – n^{2}] + b – b

= a(2n + 1)

(does not independent of n)

∴ (t_{n}) is not an arithmetic sequence.

Question 3.

If a geometric series converges which of the following is true about its common ratio r?

(i) r > 1

(ii) -1 < r < 1

(iii) r > 0

Solution:

(ii) -1 < r < 1

Question 4.

If an arithmetic series ∑tn converges, which of the following is true about t_{n}?

(i) t_{n} < 1

(ii) |t_{n}| < 1

(iii) t_{n} = 0

(iv) t_{n} → 0

Solution:

(iii) t_{n} = 0

Question 5.

Which of the following is an arithmetic-geometric series?

(i) 1 + 3x + 7x^{2} + 15x^{3}+ ….

(ii) x + \(\frac{1}{2}\)x + \(\frac{1}{3}\)x^{2} + ….

(iii) x + (1 + 2)x^{2} + (1 + 2 + 3)x^{3} +…

(iv) x + 3x^{2} + 5x^{3} + 7x^{4} + …

Solution:

(iv) x + 3x^{2} + 5x^{3} + 7x^{4} + … is an arithmetic geometric series with a = 1, d = 2, r = x.

Question 6.

For an arithmetic sequence (t_{n}) t_{p} = q, t_{q} = p, (p ≠ q), find t_{n}.

Solution:

t_{p} = q ⇒ a + (p – 1)d = q ……(1)

t_{q} = p ⇒ a + (q – 1)d = p ……(2)

From (1) and (2) we have (p – q)d = q – p

⇒ d = (-1)

Putting d = (-1) in (1)

we have a = p + q – 1

∴ t_{n} = a + (n – 1)d

= (P + q – 1) + (n – 1) (-1)

= p + q – n

Question 7.

For an arithmetic series, ∑a_{n} S_{p} = q and S_{q} = p (p ≠ q) find S_{p+q}

Solution:

S_{p} = q and S_{q} = p

Question 8.

The sum of a geometric series is 3. The series of squares of its terms have a sum of 18. Find the series.

Solution:

Question 9.

The sum of a geometric series is 14, and the series of cubes of its terms have a sum of 392 find the series.

Solution:

∴ The series is \(\sum_{n=1}^{\infty} \frac{7}{2^{n-1}}\)

Question 10.

Find the sum as directed

(i) 1 + 2a + 3a^{2} + 4a^{3} + …..(first n terms(a ≠ 1))

Solution:

(ii) 1 + (1 + x)y + (1 + x + x^{2})y^{2} + …..(to infinity)

Solution:

Let S_{∞} = 1 + (1 + x)y + (1x + x^{2})y^{2} + …

⇒ S_{n} = 1 + (1 + 1 + x)y + (1 + x + x^{2})y^{2} + ……+(1 + x + …. x^{n-1})y^{n-1}

(iii) 1 + \(\frac{3}{5}\) + \(\frac{7}{25}\) + \(\frac{15}{125}\) + \(\frac{31}{625}\) + …..(to infinity)

Solution:

(iv) 1 + 4x + 8x^{2} + 13x^{3} + 19x^{4} + …..(to infinity). Assuming that the series has a sum for |x| < 1.

Solution:

(v) 3.2 + 5.2^{2} + 7.2^{3} + …..(first n terms)

Solution:

Sn = 3.2 + 5.22 + 7.3 + ….n terms = 2[3 + 5.2 + 7.22 + ….n terms]

Question 11.

Find the sum of the infinite series.

(i) \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots\)

Solution:

= 1 – \(\frac{1}{n+1}\)

∴ \(S_{\infty}=\lim _{n \rightarrow \infty} S_n=1\)

(ii) \(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)

Solution:

\(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)

(iii) \(\frac{1}{2 \cdot 5 \cdot 8}+\frac{1}{5 \cdot 8 \cdot 11}+\frac{1}{8 \cdot 11 \cdot 14}+\ldots\)

Solution:

Here t_{n} = \(\frac{1}{(3 n-1)(3 n+2)(3 n+5)}\)

The denominator of t_{n} is the product of 3 consecutive terms of A.P. Now multiplying and dividing by (3n + 5) – (3n – 1) we have

\(t_n=\frac{(3 n+5)-(3 n-1)}{6(3 n-1)(3 n+2)(3 n+5)}\)

(iv) \(\frac{3}{1^2 \cdot 2^2}+\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\ldots\) [Hint : take t_{n} = \(\frac{2 n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\)]

Solution:

(v) \(\frac{1}{1 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{5 \cdot 9}+\ldots .\)

Solution:

Question 12.

Find S_{n} for the series.

(i) 1.2 + 2.3 + 3.4 + ….

Solution:

(ii) 1.2.3 + 2.3.4 + 3.4.5 + …

Solution:

(iii) 2.5.8 + 5.8.11 + 8.11.14 +…

Solution:

(iv) 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + …

[Hint : t_{n} = (3n – 1) (3n + 2)(3n + 5)]

Solution:

(v) 1.5 + 2.6 + 3.7 + …

[Hint: t_{n} = n(n + 4) is not a product of two successive terms of an A.P. for the term following n should be n+1, not n+4. So the method of previous exercises is not applicable. Instead, write t_{n} = n^{2} + 4n and find S_{n} = \(\sum_{k=1}^n k^2+4 \sum_{k=1}^n k\) applying formulae]

Solution:

(vi) 2.3 + 3.6 + 4.11 + …

[Hint : Take t_{n} = (n + 1) (n^{2} + 2)]

Solution:

(vii) 1.3^{2} + 2.5^{2} + 3.7^{2} + ….

Solution:

Question 13.

Find the sum of the first n terms of the series:

(i) 5 + 6 + 8 + 12 + 20 + …

Solution:

(ii) 4 + 5 + 8 + 13 + 20 + …

Solution:

Question 14.

(i) Find the sum of the product of 1,2,3….20 taken two at a time. [Hint: Required sum = \(\frac{1}{2}\left\{\left(\sum_{k=1}^{20} k\right)^2-\sum_{k=1}^{20} k^2\right\}\)]

Solution:

We know that

(x_{1} + x_{2} + x_{3} + …. + x_{n})^{2}

= (x^{2}_{1} + x^{n}_{2} + … + x^{2}_{n}) + 2 (Sum of all possible Products taken two at a time)

∴ 2 (Sum of products of 1. 2, 3,…… 20 taken two at a time)

= (1 + 2 + 3 + … 20)^{2} – (1^{2} + 2^{2} + … + 20^{2})

\(=\left(\frac{20 \times 21}{2}\right)^2-\frac{20(20+1)(40+1)}{6}\)

= (210)^{2} – 70 × 41

= 44100 – 2870 = 41230

∴ The required sum = \(\frac{41230}{2}\) = 20615

(ii) Do the same for 1, 3, 5, 7,….19.

Solution:

Here the required sum

Question 15.

If a = 1 + x + x^{2} + ….. and b = 1 + y + y^{2} + ….|x| < 1 and |y| < 1, then prove that 1 – xy + x^{2}y^{2} + x^{3}y^{3} + ….. = \(\frac{a b}{a+b-1}\)

Solution:

Question 16.

If a, b, c are respectively the pth, qth, rth terms of an A.P., then prove that a(q – r) + b(r – p) + c(p – q) = 0

Solution:

Let the first term of an A.P. = A and the common difference = D.

According to the question

t_{P} = a, t_{q} = b, t_{r} = c

⇒ A + (p – q)D = a …..(1)

A + (q – 1)D = b …..(2)

A + (r – 1)D = c …..(3)

L.H.S = a(q – r) + b (r – p) + c (p – q)

= (A + (p – 1)D) (q – r) + (A + (q – 1)D)

(r- p) + (A + (r – 1)D) (p – q)

= A (q – r + r – p + p – q) + D [(p – 1)

(q – r) + (q – 1) (r – p) + (r – 1) (p – q)]

= D (pq – pr + qr – pq + pr – qr) – D (q – r + r – p + p – q) = 0

Question 17.

If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. and a + b + c ≠ 0, prove that \(\frac{\mathbf{b}+\mathbf{c}}{\mathbf{a}}, \frac{\mathbf{c}+\mathbf{a}}{\mathbf{b}}, \frac{\mathbf{a}+\mathbf{b}}{\mathbf{c}}\) are in A.P.

Solution:

Question 18.

If a^{2}, b^{2}, c^{2} are in A.P. Prove that \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P.

Solution:

Question 19.

If \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{\mathbf{a}+\mathbf{b}}{c}\) are in A.P.,prove that \(\frac{\mathbf{1}}{\mathbf{a}}, \frac{\mathbf{1}}{\mathbf{b}}, \frac{\mathbf{1}}{\mathbf{c}}\) are inA.P.given a + b + c ≠ 0

Solution:

Question 20.

If (b – c)^{2}, (c – a)^{2}, (a – b)^{2} are in A.P., prove that \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are in A.P.

Solution:

Question 21.

If a, b, c are respectively the sum of p, q, r terms of an A.P., prove that \(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\) = 0

Solution:

Question 22.

If a, b, c,d are in G.P., prove that (a^{2} + b^{2} + c^{2})(b^{2} + c^{2} + d^{2}) = (ab + bc + cd)^{2}.

Solution:

Let a, b, c, and d are in G.P.

Let the common ratio = r

⇒ b = ar, c = ar^{2}, d = ar^{3}

LHS = (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})

= (a^{2} + a^{2}r^{2} + a^{2}r^{4}) (a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6})

= a^{4}r^{2} (1 + r^{2} + r^{4})^{2}

= (a^{2}r + a^{2}r^{3} + a^{2}r^{5})^{2}

= (a.ar + ar.ar^{2} + ar^{2}.ar^{3})^{2}

= (ab + bc + cd)^{2} = R.H.S. (proved)