Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

(I) Select the correct answer from the given alternative:

Question 1.
The common ratio for the G.P. 0.12, 0.24, 0.48, is
(A) 0.12
(B) 0.2
(C) 0.02
(D) 2
Answer:
(D) 2

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 2.
The tenth term of the geometric sequence is \(\frac{1}{4}, \frac{-1}{2}, 1,-2, \ldots\) is
(A) 1024
(B) \(\frac{1}{1024}\)
(C) -128
(D) \(\frac{-1}{128}\)
Answer:
(C) -128
Hint:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 I Q2

Question 3.
If for a G.P. \(\frac{t_{6}}{t_{3}}=\frac{1458}{54}\) then r = ?
(A) 3
(B) 2
(C) 1
(D) -1
Answer:
(A) 3
Hint:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 I Q3

Question 4.
Which term of the geometric progression 1, 2, 4, 8, ….. is 2048.
(A) 10th
(B) 11th
(C) 12th
(D) 13th
Answer:
(C) 12th
Hint:
Here, a = 1, r = 2
nth term of geometric progression = arn-1
∴ arn-1 = 2048
2n-1 = 211
n – 1 = 11
∴ n = 12

Question 5.
If the common ratio of the G.P. is 5, the 5th term is 1875, the first term is
(A) 3
(B) 5
(C) 15
(D) -5
Answer:
(A) 3

Question 6.
The sum of 3 terms of a G.P. is \(\frac{21}{4}\) and their product is 1, then the common ratio is
(A) 1
(B) 2
(C) 4
(D) 8
Answer:
(C) 4
Hint:
Let three terms be \(\frac{a}{r}\), a, ar
According to the given conditions,
\(\frac{a}{r}\) + a + ar = \(\frac{21}{4}\) …..(i)
and \(\frac{a}{r}\) × a × ar = 1,
i.e., a3 = 1
∴ a = 1
∴ from equation (i), we get
\(\frac{1}{r}\) + 1 + r = \(\frac{21}{4}\)
By solving this, we get r = 4.

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 7.
Sum to infinity of a G.P. 5, \(-\frac{5}{2}, \frac{5}{4},-\frac{5}{8}, \frac{5}{16}, \ldots\) is
(A) 5
(B) \(-\frac{1}{2}\)
(C) \(\frac{10}{3}\)
(D) \(\frac{3}{10}\)
Answer:
(C) \(\frac{10}{3}\)
Hint:
Here, a = 5, r = \(\frac{-1}{2}\), |r| < 1
∴ Sum to the infinity = \(\frac{a}{1-r}=\frac{5}{1+\frac{1}{2}}=\frac{10}{3}\)

Question 8.
The tenth term of H.P. \(\frac{2}{9}, \frac{1}{7}, \frac{2}{19}, \frac{1}{12}, \ldots\) is
(A) \(\frac{1}{27}\)
(B) \(\frac{9}{2}\)
(C) \(\frac{5}{2}\)
(D) 27
Answer:
(A) \(\frac{1}{27}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 I Q8

Question 9.
Which of the following is not true, where A, G, H are the AM, GM, HM of a and b respectively, (a, b > 0)
(A) A = \(\frac{a+b}{2}\)
(B) G = \(\sqrt{a b}\)
(C) H = \(\frac{2 a b}{a+b}\)
(D) A = GH
Answer:
(D) A = GH

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 10.
The G.M. of two numbers exceeds their H.M. by \(\frac{6}{5}\), the A.M. exceeds G.M. by \(\frac{3}{2}\) the two numbers are
(A) 6, \(\frac{15}{2}\)
(B) 15, 25
(C) 3, 12
(D) \(\frac{6}{5}\), \(\frac{3}{2}\)
Answer:
(C) 3, 12
Hint:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 I Q10
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 I Q10.1

(II) Answer the following:

Question 1.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q1

Question 2.
Find the sum of the first 5 terms of the G.P. whose first term is 1 and the common ratio is \(\frac{2}{3}\).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q2

Question 3.
For a G.P. a = \(\frac{4}{3}\) and t7 = \(\frac{243}{1024}\), find the value of r.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q3

Question 4.
For a sequence, if \(t_{n}=\frac{5^{n-2}}{7^{n-3}}\), verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Solution:
The sequence (tn) is a G.P.
if \(\frac{\mathrm{t}_{\mathrm{n}+1}}{\mathrm{t}_{\mathrm{n}}}\) = constant for all n ∈ N.
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q4

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 5.
Find three numbers in G.P. such that their sum is 35 and their product is 1000.
Solution:
Let the three numbers in G.P. be \(\frac{a}{r}\), a, ar.
According to the given conditions,
\(\frac{a}{r}\) + a + ar = 35
a(\(\frac{1}{r}\) + 1 + r) = 35 …..(i)
Also, (\(\frac{a}{r}\))(a)(ar) = 1000
a3 = 1000
∴ a = 10
Substituting the value of a in (i), we get
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q5
Hence, the three numbers in G.P. are 20, 10, 5, or 5, 10, 20.

Question 6.
Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.
Solution:
Let the five numbers in G.P. be \(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\).
According to the given condition,
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q6
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q6.1

Question 7.
For a sequence, Sn = 4(7n – 1), verify that the sequence is a G.P.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q7
∴ The given sequence is a G.P.

Question 8.
Find 2 + 22 + 222 + 2222 + … upto n terms.
Solution:
Sn = 2 + 22 + 222 +… upto n terms
= 2(1 + 11 + 111 + ….. upto n terms)
= \(\frac{2}{9}\) (9 + 99 + 999 + … upto n terms)
= \(\frac{2}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) + …… upto n terms]
= \(\frac{2}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)]
Since 10, 100, 1000, ….. n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10,
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q8

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 9.
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,…
Solution:
0.6, 0.66, 0.666, 0.6666, …
∴ t1 = 0.6
t2 = 0.66 = 0.6 + 0.06
t3 = 0.666 = 0.6 + 0.06 + 0.006
Hence, in general
tn = 0.6 + 0.06 + 0.006 + …..upto n terms.
The terms are in G.P. with
a = 0.6, r = \(\frac{0.06}{0.6}\) = 0.1
∴ tn = the sum of first n terms of the G.P.
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q9

Question 10.
Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q10

Question 11.
Find \(\sum_{r=1}^{n} r(r-3)(r-2)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q11

Question 12.
Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\)
Solution:
We know that
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q12

Question 13.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots r^{3}}{(r+1)^{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q13

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 14.
Find 2 × 6 + 4 × 9 + 6 × 12 + ….. upto n terms.
Solution:
2, 4, 6, ….. are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
6, 9, 12, ….. are in A.P.
∴ rth term = 6 + (r – 1)(3) = (3r + 3)
∴ 2 × 6 + 4 × 9 + 6 × 12 + ….. to n terms
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q14
= n(n + 1) [2n + 1 + 3]
= 2n(n + 1)(n + 2)

Question 15.
Find 2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + …… upto n terms.
Solution:
2, 4, 6,… are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
5, 7, 9, … are in A.P.
∴ rth term = 5 + (r – 1) (2) = (2r + 3)
8, 10, 12, … are in A.P.
∴ rth term = 8 + (r – 1) (2) = (2r + 6)
2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + ….. to n terms
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q15
= 2n (n + 1) [n(n + 1) + 3(2n + 1) + 9]
= 2n (n + 1)(n2 + 7n + 12)
= 2n (n + 1) (n + 3) (n + 4)

Question 16.
Find \(\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{2}+\frac{1^{2}+2^{2}+3^{2}}{3}+\ldots\) upto n terms.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q16
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q16.1

Question 17.
Find 122 + 132 + 142 + 152 + ….. 202
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q17

Question 18.
If \(\frac{1+2+3+4+5+\ldots \text { upto } \mathrm{n} \text { terms }}{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \text { upto } \mathrm{n} \text { terms }}=\frac{3}{22}\), Find the value of n.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q18

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 19.
Find (502 – 492) + (482 – 472) + (462 – 452) +… + (22 – 12).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q19

Question 20.
If \(\frac{1 \times 3+2 \times 5+3 \times 7+\ldots \text { upto } \mathrm{n} \text { terms }}{1^{3}+2^{3}+3^{3}+\ldots \text { upto } \mathrm{n} \text { terms }}=\frac{5}{9}\), find the value of n.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q20
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q20.1

Question 21.
For a G.P. if t2 = 7, t4 = 1575, find a.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q21

Question 22.
If for a G.P. t3 = \(\frac{1}{3}\), t6 = \(\frac{1}{81}\) find r.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q22

Question 23.
Find \(\sum_{r=1}^{n}\left(\frac{2}{3}\right)^{r}\).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q23

Question 24.
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
Solution:
Since k – 1, k, k + 2 are consecutive terms of a G.P.,
\(\frac{k}{k-1}=\frac{k+2}{k}\)
k2 = k2 + k – 2
k – 2 = 0
∴ k = 2

Question 25.
If for a G.P. first term is (27)2 and the seventh term is (8)2, find S8.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q25

Question 26.
If pth, qth and rth terms of a G.P. are x, y, z respectively. Find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Solution:
Let a be the first term and R be the common ratio of the G.P.
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q26

Question 27.
Which 2 terms are inserted between 5 and 40 so that the resulting sequence is G.P.
Solution:
Let the required numbers be G1 and G2.
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q27
∴ For the resulting sequence to be in G.P. we need to insert numbers 10 and 20.

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 28.
If p, q, r are in G.P. and \(\mathrm{p}^{1 / \mathrm{x}}=\mathrm{q}^{1 / \mathrm{y}}=\mathrm{r}^{1 / \mathrm{z}}\), verify whether x, y, z are in A.P. or G.P. or neither.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q28

Question 29.
If a, b, c are in G.P. and ax2 + 2bx + c = 0 and px2 + 2qx + r = 0 have common roots, then verify that pb2 – 2qba + ra2 = 0.
Solution:
a, b, c are in G.P.
∴ b2 = ac
ax2 + 2bx + c = 0 becomes
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q29

Question 30.
If p, q, r, s are in G.P., show that (p2 + q2 + r2)(q2 + r2 + s2) = (pq + qr + rs)2.
Solution:
p, q, r, s are in G.P.
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q30

Question 31.
If p, q, r, s are in G.P., show that (pn + qn), (qn + rn), (rn + sn) are also in G.P.
Solution:
p, q, r, s are in G.P.
Let the common ratio be R
∴ let p = \(\frac{a}{R^{3}}\), q = \(\frac{a}{R}\), r = aR and s = aR3
To show that (pn + qn), (qn + rn), (rn + sn) are in G.P,
i.e., we have to show
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q31

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2

Question 32.
Find the coefficient x6 in the expression of e2x using series expansion.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q32

Question 33.
Find the sum of infinite terms of \(1+\frac{4}{5}+\frac{7}{25}+\frac{10}{125}+\frac{13}{625}+\ldots\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q33
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Miscellaneous Exercise 2 II Q33.1