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

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4

Question 1.

In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.

Solution:

Question 2.

For a G.P. a = \(\frac{4}{3}\) and t_{7} = \(\frac{243}{1024}\), find the value of r.

Solution:

Question 3.

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 (t_{n}) is a G.P., if \(\frac{5^{n-2}}{7^{n-3}}\) = constant, for all n ∈ N.

∴ the sequence is a G.P. with common ratio = \(\frac{5}{7}\)

∴ first term = t_{1} = \(\frac{5^{1-2}}{7^{1-3}}=\frac{5^{-1}}{7^{-2}}=\frac{7^{2}}{5}=\frac{49}{5}\)

Question 4.

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 first condition,

∴ the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.

Question 5.

Find 4 numbers in G. P. such that the sum of the middle 2 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 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 first condition,

Question 7.

For a sequence, S_{n} = 4(7^{n} – 1), verify whether the sequence is a G.P.

Solution:

Question 8.

Find 2 + 22 + 222 + 2222 + …… upto n terms.

Solution:

S_{n} = 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

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, ……

∴ t_{1} = 0.6

t_{2} = 0.66 = 0.6 + 0.06

t_{3} = 0.666 = 0.6 + 0.06 + 0.006

Hence, in general

t_{n} = 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

∴ t_{n} = the sum of first n terms of the G.P.

Question 10.

Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\).

Solution:

Question 11.

Find \(\sum_{\mathbf{r}=1}^{\mathbf{n}} \mathbf{r}(\mathbf{r}-\mathbf{3})(\mathbf{r}-\mathbf{2})\).

Solution:

Question 12.

Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\)

Solution:

We know that,

Question 13.

Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots+r^{3}}{(r+1)^{2}}\)

Solution:

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 +…… upto n terms

= n(n + 1) (2n + 1 + 3)

= 2n(n + 1)(n + 2)

Question 15.

Find 12^{2} + 13^{2} + 14^{2} + 15^{2} + …… + 20^{2}.

Solution:

12^{2} + 13^{2} + 14^{2} + 15^{2} + …… + 20^{2}

= (1^{2} + 2^{2} + 3^{2} + 4^{2} + ……. + 202) – (1^{2} + 2^{2} + 3^{2} + 4^{2} + …… + 11^{2})

= 2870 – 506

= 2364

Question 16.

Find (50^{2} – 49^{2}) + (48^{2} – 47^{2}) + (46^{2} – 45^{2}) + …… + (2^{2} – 1^{2}).

Solution:

(50^{2} – 49^{2}) + (48^{2} – 47^{2}) + (46^{2} – 45^{2}) + …… + (2^{2} – 1^{2})

= (50^{2} + 48^{2} + 46^{2} + …… + 2^{2}) – (49^{2} + 47^{2} + 45^{2} + …… + 1^{2})

= \(\sum_{r=1}^{25}(2 r)^{2}-\sum_{r=1}^{25}(2 r-1)^{2}\)

= 1300 – 25

= 1275

Question 17.

In a G.P., if t_{2} = 7, t_{4} = 1575, find r.

Solution:

Question 18.

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.

∴ \(\frac{k}{k-1}=\frac{k+2}{k}\)

∴ k^{2} = k^{2} + k – 2

∴ k – 2 = 0

∴ k = 2

Question 19.

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.

∴ t_{n} = \(\text { a. } R^{n-1}\)

∴ x = \(\text { a. } R^{p-1}\), y = \(\text { a. } R^{q-1}\), z = \(\text { a. } R^{r-1}\)