Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Question 1.
Verify whether the following sequences are G.P. If so, write tn.
(i) 2, 6, 18, 54, ……
(ii) 1, -5, 25, -125, …….
(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots\)
(iv) 3, 4, 5, 6, ……
(v) 7, 14, 21, 28, …..
Solution:
(i) 2, 6, 18, 54, …….
t1 = 2, t2 = 6, t3 = 18, t4 = 54, …..
Here, \(\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}=\frac{t_{4}}{t_{3}}=3\)
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, a = 2, r = 3
tn= arn-1
∴ tn = 2(3n-1)

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

(ii) 1, -5, 25, -125, ……
t1 = 1, t2 = -5, t3 = 25, t4 = -125, …..
Here, \(\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}=\frac{t_{4}}{t_{3}}=-5\)
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, a = 1, r = -5
tn = arn-1
∴ tn = (-5)n-1

(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q1
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q1.1

(iv) 3, 4, 5, 6,……
t1 = 3, t2 = 4, t3 = 5, t4 = 6, …..
Here, \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{4}{3}, \frac{\mathrm{t}_{3}}{\mathrm{t}_{2}}=\frac{5}{4}, \frac{\mathrm{t}_{4}}{\mathrm{t}_{3}}=\frac{6}{5}\)
Since, \(\frac{t_{2}}{t_{1}} \neq \frac{t_{3}}{t_{2}} \neq \frac{t_{4}}{t_{3}}\)
∴ the given sequence is not a geometric progression.

(v) 7, 14, 21, 28, …..
t1 = 7, t2 = 14, t3 = 21, t4 = 28, …..
Here, \(\frac{t_{2}}{t_{1}}=2, \frac{t_{3}}{t_{2}}=\frac{3}{2}, \frac{t_{4}}{t_{3}}=\frac{4}{3}\)
Since, \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}} \neq \frac{\mathrm{t}_{3}}{\mathrm{t}_{2}} \neq \frac{\mathrm{t}_{4}}{\mathrm{t}_{3}}\)
∴ the given sequence is not a geometric progression.

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Question 2.
For the G.P.,
(i) if r = \(\frac{1}{3}\), a = 9, find t7.
(ii) if a = \(\frac{7}{243}\), r = \(\frac{1}{3}\), find t3.
(iii) if a = 7, r = -3, find t6.
(iv) if a = \(\frac{2}{3}\), t6 = 162, find r.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q2
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q2.1

Question 3.
Which term of the G. P. 5, 25, 125, 625, ….. is 510?
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q3

Question 4.
For what values of x, \(\frac{4}{3}\), x, \(\frac{4}{27}\) are in G. P.?
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q4

Question 5.
If for a sequence, \(t_{n}=\frac{5^{n-3}}{2^{n-3}}\), show that the sequence is a G. P. Find its first term and the common ratio.
Solution:
The sequence (tn) is a G.P., if \(\frac{t_{n}}{t_{n-1}}\) = constant, for all n ∈ N
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q5
∴ the sequence is a G. P. with common ratio \(\frac{5}{2}\)
First term, t1 = \(\frac{5^{\mathrm{l}-3}}{2^{1-3}}=\frac{2^{2}}{5^{2}}=\frac{4}{25}\)

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Question 6.
Find three numbers in G. P. such that their sum is 21 and sum of their squares is 189.
Solution:
Let the three numbers in G. P. be \(\frac{a}{\mathrm{r}}\), a, ar.
According to the first condition,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q6
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q6.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q6.2
∴ the three numbers are 12, 6, 3 or 3, 6, 12.
Check:
First condition:
12, 6, 3 are in G.P. with r = \(\frac{1}{2}\)
12 + 6 + 3 = 21
Second condition:
122 + 62 + 32 = 144 + 36 + 9 = 189
Thus, both the conditions are satisfied.

Question 7.
Find four numbers in G. P. such that sum of the middle two numbers is \(\frac{10}{3}\) and their product is 1.
Solution:
Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\).
According to the second condition,
\(\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1\)
∴ a4 = 1
∴ a = 1
According to the first condition,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q7

Question 8.
Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term.
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 conditions,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q8
When a = 4, r = -2
\(\frac{a}{r^{2}}\) = 1, \(\frac{a}{r}\) = -2, a = 4, ar = -8, ar2 = 16
∴ the five numbers in G.P. are 1, 2, 4, 8, 16 or 1, -2, 4, -8, 16.

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Question 9.
The fifth term of a G. P. is x, eighth term of the G. P. is y and eleventh term of the G. P. is z. Verify whether y2 = xz.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q9

Question 10.
If p, q, r, s are in G. P., show that p + q, q + r, r + s are also in G.P.
Solution:
p, q, r, s are in G.P.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q10
∴ p + q, q + r, r + s are in G.P.