Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.4 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.4

Question 1.

Verify whether the following sequences are H.P.

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

Solution:

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

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

t_{1} = 3, t_{2} = 5, t_{3} = 7, t_{4} = 9, …..

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 a H.P.

(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots\)

Solution:

\(\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots\)

Here, the reciprocal sequence is 3, 6, 12, 24,…

t_{1} = 3, t_{2} = 6, t_{3} = 12, ……

t_{2} – t_{1} = 3, t_{3} – t_{2} = 6

t_{2} – t_{1} ≠ t_{3} – t_{2}

∴ The reciprocal sequence is not an A.P.

∴ The given sequence is not a H.P.

(iii) \(5, \frac{10}{17}, \frac{10}{32}, \frac{10}{47}, \ldots\)

Solution:

∴ The reciprocal sequence is an A.P.

∴ The given sequence is a H.P.

Question 2.

Find the nth term and hence find the 8th term of the following HPs.

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

Solution:

\(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots\) are in H.P.

∴ 2, 5, 8, 11,… are in A.P.

∴ a = 2, d = 3

t_{n} = a + (n – 1)d

= 2 + (n – 1)(3)

= 3n – 1

∴ nth term of H.P. = \(\frac{1}{3 n-1}\)

∴ 8th term of H.P. = \(\frac{1}{3(8)-1}\) = \(\frac{1}{23}\)

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

Solution:

\(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots\) are in H.P.

∴ 4, 6, 8, 10, … are in A.P.

∴ a = 4, d = 2

t_{n} = a + (n – 1)d

= 4 + (n – 1) (2)

= 2n + 2

∴ nth term of H.P. = \(\frac{1}{2 n+2}\)

∴ 8th term of H.P. = \(\frac{1}{2(8)+2}\) = \(\frac{1}{18}\)

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

Solution:

\(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \ldots\) are in H.P.

∴ 5, 10, 15, 20, … are in A.P.

∴ a = 5, d = 5

t_{n} = a + (n – 1)d

= 5 + (n – 1) (5)

= 5n

∴ nth term of H.P. = \(\frac{1}{5 n}\)

∴ 8th term of H.P. = \(\frac{1}{5(8)}\) = \(\frac{1}{40}\)

Question 3.

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

Solution:

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

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

∴ 4^{2} = A.M. × \(\frac{16}{5}\)

∴ A.M. = 16 × \(\frac{5}{16}\)

∴ 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

Now, (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 \(\frac{1}{4}\) and \(\frac{1}{3}\) so that the resulting sequence is a H.P.

Solution:

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

∴ \(\frac{1}{4}, \frac{1}{\mathrm{H}_{1}}, \frac{1}{\mathrm{H}_{2}}, \frac{1}{3}\) are in H.P.

∴ 4, H_{1}, H_{2}, 3 are in A.P.

t_{1} = 4, t_{2} = H_{1}, t_{3} = H_{2}, t_{4} = 3

∴ t_{1} = a = 4, t_{4} = 3

t_{n} = a + (n – 1)d

t_{4} = 4 + (4 – 1)d

3 = 4 + 3d

3d = -1

∴ d = \(\frac{-1}{3}\)

H_{1} = t_{2} = a + d = 4 – \(\frac{1}{3}\) = \(\frac{11}{3}\)

H_{2} = t_{3} = a + 2d = 4 – \(\frac{2}{3}\) = \(\frac{10}{3}\)

∴ For resulting sequence to be H.P. we need to insert numbers \(\frac{3}{11}\) and \(\frac{3}{10}\).

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^{2} = 1(-3)^{2} = 9

∴ For resulting sequence to be G.P. we need to insert numbers -3 and 9.

Question 8.

If the A.M. of two numbers exceeds their G.M. by 2 and their H.M. by \(\frac{18}{5}\), find the numbers.

Solution:

Let a and b be the two numbers.

A = \(\frac{a+b}{2}\), G = \(\sqrt{a b}\), H = \(\frac{2 a b}{a+b}\)

According to the given conditions,

Consider, G = A – 2 = 10 – 2 = 8

\(\sqrt{a b}\) = 8

ab = 64

a(20 – a) = 64 …..[From (i)]

a^{2} – 20a + 64 = 0

(a – 4)(a – 16) = 0

∴ a = 4 or a = 16

When a = 4, b = 20 – 4 = 16

When a = 16, b = 20 – 16 = 4

∴ The two numbers are 4 and 16.

Question 9.

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 and b be the two numbers.

A = \(\frac{a+b}{2}\), G = \(\sqrt{a b}\), H = \(\frac{2 a b}{a+b}\)

According to the given conditions,

\(\sqrt{a b}\) = 6

ab = 36

a(13 – a) = 36 ……[From (i)]

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.