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Answer 1

Questions:-

1. Find the fourth term of the sequence 2 , 2 1/2 , 3 1/3 ....
2. Insert two harmonic means between 5 & 11.
3. Insert four harmonic means between 2/3 and 2/13.
4. If 12 and 9 3/5 are the geometric and harmonic means, respectively, between two numbers, find them.

Answers:-

1) Given sequence;

2 , 2 1/2 , 3 1/3...

i.e., 2 , 5/2 , 10/3 , .... is in HP

Therefore;

1/2 , 2/5 , 3/10 ... is in AP.

Here;

• a = 1/2
• d = 2/5 - 1/2 = (4 - 5)/10 = - 1/10

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

⟹ a₄ = 1/2 + (4 - 1)(- 1/10)

⟹ a₄ = 1/2 - 3/10

⟹ a₄ = (5 - 3)/10

⟹ a₄ = 2/10

⟹ a₄ = 1/5

1/5 is the 4th term in AP.

∴ 5 is the 4th term in the given HP.

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2) Given numbers are 5 and 11.

We know that,

Harmonic mean between a , b (HM) = 2ab / (a + b)

So,

⟹ HM₁ = 2(5)(11)/5 + 11

⟹ HM₁ = 110/16

⟹ HM₁ = 55/8

Now,

5 , 55/8 , 11 are in HP.

⟹ HM₂ = HM of 5 , 55/8.

$\implies \sf \: HM_2 = \dfrac{2(5) \big( \frac{55}{8} \big) }{ 5 + \frac{55}{8} } \\ \\ \\ \implies \sf \: HM_2 = \frac{55 \times 5}{4} \times \frac{8}{40 + 55} \\ \\ \\ \implies \sf \:HM_2 = \frac{55 \times 5}{4} \times \frac{8}{95} \\ \\ \\ \implies \boxed{\sf \:HM_2 = \frac{110}{19} }$

∴ 55/8 , 110/19 are the two harmonic means between 5 & 11.

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3) Given numbers are 2/3 and 2/13.

We know;

HM = 2ab / a + b

So,

$\sf \bigstar \: HM_1 = \frac{2 \times \frac{2}{3} \times \frac{2}{13} }{ \frac{2}{3} + \frac{2}{13} } \\ \\ \\ \implies \sf \: HM_1 = \frac{8}{39} \times \frac{39}{26 + 6} \\ \\ \\ \implies \sf \:HM_1 = \frac{8}{32} \\ \\ \\ \implies \boxed{ \sf \: HM_1 = \frac{1}{4} }$

Now,

2/3 , 1/4 , 2/13 are in HP.

⟹ HM₂ = HM of 2/3 and 1/4.

$\implies \sf \: HM_2 = \frac{2 \times \frac{2}{3} \times \frac{1}{4} }{ \frac{2}{3} + \frac{1}{4} } \\ \\ \\ \implies \sf \:HM_2 = \frac{1}{3} \times \frac{12}{8 + 3} \\ \\ \\ \implies \boxed{\sf \:HM_2 = \frac{4}{11} }$

Again;

2/3 , 4/11 , 1/4 , 2/3.... are in HP.

⟹ HM₃ = HM of 4/11 & 1/4.

$\: \implies \sf \:HM_3 = \frac{2 \times \frac{4}{11} \times \frac{1}{4} }{ \frac{4}{11} + \frac{1}{4} } \\ \\ \\ \: \implies \sf \:HM_3 = \frac{2}{11} \times \frac{44}{16 + 11} \\ \\ \\ \: \implies \sf \:HM_3 = \frac{2 \times 4}{27} \\ \\ \\ \: \implies \boxed{\sf \:HM_3 = \frac{8}{27} }$

Similarly;

⟹ HM₄ = HM of 8/27 & 1/4.

$\: \: \implies \sf \:HM_4 = \frac{2 \times \frac{8}{27} \times \frac{1}{4} }{ \frac{8}{27} + \frac{1}{4} } \\ \\ \\ \: \implies \sf \:HM_4 = \frac{4}{27} \times \frac{108}{32 + 27} \\ \\ \\ \: \implies \sf \:HM_4 = \frac{4 \times 4}{59} \\ \\ \\ \: \implies \boxed{\sf \:HM_4 = \frac{16}{59} }$

∴ 1/4 , 4/11 , 8/27 , 16/59 are the four harmonic means between 2/3 & 2/13.

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4) Let the two numbers be a and b.

Given:-

Geometric mean = 12

We know that,

GM of a , b = √ab

So,

⟹ √ab = 12

On squaring both sides we get,

⟹ (√ab)² = 144

⟹ ab = 144 -- equation (1)

Also given that,

Harmonic mean = 9 3/5 = 48/5.

We know,

HM = 2ab / (a + b)

So,

$\sf \: \frac{2ab}{a + b} = \frac{48}{5}$

Substitute the value of ab from equation (1).

$\implies \sf \: \frac{2 \times 144}{a + b} = \frac{48}{5} \\ \\ \\ \implies \sf \:2 \times 144 \times 5 = (a + b) \times 48 \\ \\ \\ \implies \sf \: \frac{2 \times 144 \times 5}{48} = a + b \\ \\ \\ \implies \sf \:30 = a + b \: \: - - \: \: equation \: (2).$

We know that,

(a - b)² = (a + b)² - 4ab

Putting the respective values we get;

⟹ (a - b)² = (30)² - 4(144)

⟹ (a - b)² = 900 - 576

⟹ (a - b)² = 324

⟹ (a - b) = √324

⟹ a - b = 18 -- equation (3)

Add equations (2) & (3).

⟹ a + b + a - b = 30 + 18

⟹ 2a = 48

⟹ a = 48/2

⟹ a = 24

Substitute the value of a in equation (2).

⟹ a + b = 30

⟹ 24 + b = 30

⟹ b = 30 - 24

⟹ b = 6

∴ The required numbers are 24,6.