Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.4 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4

Question 1.

Verify whether the following sequences are H.P.

(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)

(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \ldots \ldots \ldots \ldots\)

(iii) \(\frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \ldots\)

Solution:

(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)

Here, the reciprocal sequence is 3, 5, 7, 9, …

∴ t_{1} = 3, t_{2} = 5, t_{3} = 7, …..

∵ t_{2} – t_{1} = t_{3} – t_{2} = t_{4} – t_{3} = 2, constant

∴ The reciprocal sequence is an A.P.

∴ the given sequence is H.P.

(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \ldots \ldots \ldots \ldots\)

Here, the reciprocal sequence is 3, 6, 9, 12 …

∴ t_{1} = 3, t_{2} = 6, t_{3} = 9, t_{4} = 12, …

∵ t_{2} – t_{1} = t_{3} – t_{2} = t_{4} – t_{3} = 3, constant

∴ The reciprocal sequence is an A.P.

∴ The given sequence is H.P.

(iii) \(\frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \ldots\)

Here, the reciprocal sequence is 7, 9, 11, 13, 15, ……

∴ t_{1} = 7, t_{2} = 9, t_{3} = 11, t_{4} = 13, …..

∵ t_{2} – t_{1} = t_{3} – t_{2} = t_{4} – t_{3} = 2, constant

∴ The reciprocal sequence is an A.P.

∴ The given sequence is H.P.

Question 2.

Find the nth term and hence find the 8th term of the following H.P.s:

(i) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots \ldots \ldots\)

(ii) \(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots \ldots \ldots \ldots\)

(iii) \(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \cdots \cdots \cdots\)

Solution:

Question 3.

Find A.M. of two positive numbers whose G.M. and H.M. are 4 and \(\frac{16}{5}\).

Solution:

G.M. = 4, H.M. = \(\frac{16}{5}\)

∵ (G.M.)^{2} = (A.M.) (H.M.)

∴ 16 = A.M. × \(\frac{16}{5}\)

∴ A.M. = 5

Question 4.

Find H.M. of two positive numbers whose A.M. and G.M. are \(\frac{15}{2}\) and 6.

Solution:

A.M. = \(\frac{15}{2}\), G.M. = 6

Now, (G.M.)^{2} = (A.M.) (H.M.)

∴ 6^{2} = \(\frac{15}{2}\) × H.M.

∴ H.M. = 36 × \(\frac{2}{15}\)

∴ H.M. = \(\frac{24}{5}\)

Question 5.

Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48.

Solution:

A.M. = 75, H.M. = 48

(G.M.)^{2} = (A.M.) (H.M.)

∵ (G.M.)^{2} = 75 × 48

∵ (G.M.)^{2} = 25 × 3 × 16 × 3

∵ (G.M.)^{2} = 5^{2} × 4^{2} × 3^{2}

∴ G.M. = 5 × 4 × 3

∴ G.M. = 60

Question 6.

Insert two numbers between \(\frac{1}{7}\) and \(\frac{1}{13}\) so that the resulting sequence is a H.P.

Solution:

Let the required numbers be \(\frac{1}{\mathrm{H}_{1}}\) and \(\frac{1}{\mathrm{H}_{2}}\).

∴ \(\frac{1}{7}, \frac{1}{\mathrm{H}_{1}}, \frac{1}{\mathrm{H}_{2}}, \frac{1}{13}\) are in H.P.

∴ 7, H_{1}, H_{2} and 13 are in A.P.

∴ t_{1} = a = 7 and t_{4} = a + 3d = 13

∴ 7 + 3d = 13

∴ 3d = 6

∴ d = 2

∴ H_{1} = t_{2} = a + d = 7 + 2 = 9

and H_{2} = t_{3} = a + 2d = 7 + 2(2) = 11

∴ \(\frac{1}{9}\) and \(\frac{1}{11}\) are the required numbers to be inserted between \(\frac{1}{7}\) and \(\frac{1}{13}\) so that the resulting sequence is a H.P.

Question 7.

Insert two numbers between 1 and -27 so that the resulting sequence is a G.P.

Solution:

Let the required numbers be G_{1} and G_{2}.

∴ 1, G_{1}, G_{2}, -27 are in G.P.

∴ t_{1} = 1, t_{2} = G_{1}, t_{3} = G_{2}, t_{4} = -27

∴ t_{1} = a = 1

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

∴ t_{4} = (1) r^{4-1}

∴ -27 = r^{3}

∴ r^{3} = (-3)^{3}

∴ r = -3

∴ G_{1} = t_{2} = ar = 1(-3) = -3

∴ G_{2} = t_{3} = ar = 1(-3)^{2} = 9

∴ -3 and 9 are the required numbers to be inserted between 1 and -27 so that the resulting sequence is a G.P.

Question 8.

Find two numbers whose A.M. exceeds their G.M. by \(\frac{1}{2}\) and their H.M. by \(\frac{25}{26}\).

Solution:

Let a, b be the two numbers.

∴ a + b = 13

∴ b = 13 – a …….(iii)

and ab = 36

∴ a(13 – a) = 36 …… [From (iii)]

∴ a^{2} – 13a + 36 = 0

∴ (a – 4)(a – 9) = 0

∴ a = 4 or a = 9

When a = 4, b = 13 – 4 = 9

When a = 9, b = 13 – 9 = 4

∴ the two numbers are 4 and 9.

Question 9.

Find two numbers whose A.M. exceeds G.M. bv 7 and their H.M. by \(\frac{63}{5}\).

Solution:

Let a, b be the two numbers.

∴ a + b = 70

∴ b = 70 – a …..(ii)

∴ G = A – 7 = 35 – 7 = 28 …….[From (i)]

∴ √ab = 28

∴ ab = 28^{2} = 784

∴ a(70 – a) = 784 ……[From (ii)]

∴ 70a – a^{2} = 784

∴ a^{2} – 70a + 784 = 0

∴ a^{2} – 56a – 14a + 784 = 0

∴ (a – 56) (a – 14) = 0

∴ a = 14 or a = 56

When a = 14, b = 70 – 14 = 56

When a = 56, b = 70 – 56 = 14

∴ the two numbers are 14 and 56.