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, …….

t_{1} = 2, t_{2} = 6, t_{3} = 18, t_{4} = 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

t_{n}= ar^{n-1}

∴ t_{n} = 2(3^{n-1})

(ii) 1, -5, 25, -125, ……

t_{1} = 1, t_{2} = -5, t_{3} = 25, t_{4} = -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

t_{n} = ar^{n-1}

∴ t_{n} = (-5)^{n-1}

(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots\)

Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.

(iv) 3, 4, 5, 6,……

t_{1} = 3, t_{2} = 4, t_{3} = 5, t_{4} = 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, …..

t_{1} = 7, t_{2} = 14, t_{3} = 21, t_{4} = 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.

Question 2.

For the G.P.,

(i) if r = \(\frac{1}{3}\), a = 9, find t_{7}.

(ii) if a = \(\frac{7}{243}\), r = \(\frac{1}{3}\), find t_{3}.

(iii) if a = 7, r = -3, find t_{6}.

(iv) if a = \(\frac{2}{3}\), t_{6} = 162, find r.

Solution:

Question 3.

Which term of the G. P. 5, 25, 125, 625, ….. is 510?

Solution:

Question 4.

For what values of x, \(\frac{4}{3}\), x, \(\frac{4}{27}\) are in G. P.?

Solution:

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

∴ the sequence is a G. P. with common ratio \(\frac{5}{2}\)

First term, t_{1} = \(\frac{5^{\mathrm{l}-3}}{2^{1-3}}=\frac{2^{2}}{5^{2}}=\frac{4}{25}\)

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,

∴ 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:

12^{2} + 6^{2} + 3^{2} = 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,

\(\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1\)

∴ a^{4} = 1

∴ a = 1

According to the first condition,

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,

When a = 4, r = -2

\(\frac{a}{r^{2}}\) = 1, \(\frac{a}{r}\) = -2, a = 4, ar = -8, ar^{2} = 16

∴ the five numbers in G.P. are 1, 2, 4, 8, 16 or 1, -2, 4, -8, 16.

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 y^{2} = xz.

Solution:

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.

∴ p + q, q + r, r + s are in G.P.