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T_p of H.P = qr ⇒ T_p of A.P. = \(\frac{1}{qr}\) ⇒ a + (p - 1)d = \(\frac{1}{qr}\)

Where a and d are the first term and common difference respectively of the A.P

Also T_q of H.P. = pr ⇒ T_q of A.P = \(\frac{1}{pr}\) ⇒ a + (q - 1)d = \(\frac{1}{pr}\)

Eqn (ii) – Eqn (i) ⇒ (q – 1)d – (p – 1)d = \(\frac{1}{pr}\) - \(\frac{1}{qr}\) ⇒ (q – p)d = \(\frac{q-p}{pqr}\) ⇒ d = \(\frac{1}{pqr}\)

⇒ a + (p – 1)\(\frac{1}{pqr}\) = \(\frac{1}{qr}\) ⇒ a = \(\frac{1}{qr}\) - (p-1)\(\frac{1}{pqr}\) = \(\frac{1}{qr}\) - \(\frac{1}{qr}\) + \(\frac{1}{pqr}\) = \(\frac{1}{pqr}\)

∴ T_r of A.P = a + (r - 1)d = \(\frac{1}{pqr}\) + (r - 1)\(\frac{1}{pqr}\) = \(\frac{1}{pqr}\) + \(\frac{1}{pq}\) - \(\frac{1}{pqr}\) = \(\frac{1}{pq}\)

⇒ T_r of A.P = pq.

H being the H.M. between x and y

⇒ H = \(\frac{2xy}{x+y}\) ⇒ \(\frac{H}{x}\) = \(\frac{2y}{x+y}\) and \(\frac{H}{y}\) = \(\frac{2x}{x+y}\)

⇒ \(\frac{H+x}{H-x}\) = \(\frac{2y+x+y}{2y-(x+y)}\) and \(\frac{H+y}{H-y}\) = \(\frac{2x+x+y}{2x-(x+y)}\) 

(Using Componendo and Dividendo)

⇒ \(\frac{H+x}{H-x}\) = \(\frac{3y+x}{y-x}\) and \(\frac{H+y}{H-y}\) = \(\frac{3x+y}{x-y}\)

∴ \(\frac{H+x}{H-x}\) + \(\frac{H+y}{H-y}\) = \(\frac{3y+x}{y-x}\) + \(\frac{3x+y}{x-y}\) = \(\frac{3y+x-3x-y}{y-x}\) = \(\frac{2(y-x)}{y-x}\) = 2.

3 harmonic means between 5 and 6

⇒ 3 arithmetic means between \(\frac{1}{5}\) and \(\frac{1}{6}\).

Let A₁, A₂, A₃ be the arithmetic means between \(\frac{1}{5}\) and \(\frac{1}{6}\).

Then, \(\frac{1}{5}\), A₁ , A₂ A₃, \(\frac{1}{6}\) from an A.P,

Where t₁ = a = \(\frac{1}{5}\), t₅ = a + 4d = \(\frac{1}{6}\)

∴ \(\frac{1}{5}\)+ 4d = \(\frac{1}{6}\) ⇒ 4d = \(\frac{1}{6}\) - \(\frac{1}{5}\) = \(-\frac{1}{30}\) ⇒ d = \(-\frac{1}{120}\)

∴ A₁ = a + d = \(\frac{1}{5}\) + \(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{23}{120}\)

A₂ = a + 2d = \(\frac{1}{5}\) + 2 x\(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{1}{5}\) - \(\frac{1}{60}\) = \(\frac{11}{60}\)

A₃ = a + 3d = \(\frac{1}{5}\) + 3 x\(\bigg(-\frac{1}{120}\bigg)\) = \(\frac{1}{5}\) - \(\frac{1}{40}\) = \(\frac{7}{40}\)

∴ Required harmonic means are \(\frac{120}{23}\),\(\frac{60}{11}\), \(\frac{40}{7}\).

Let the roots of the equation be α and β. Then, 

Sum of roots = α + β = \(\frac{(4+\sqrt5)}{5+\sqrt2}\)

Product of roots = αβ = \(\frac{8+2\sqrt5}{5+\sqrt2}\)

Now, Harmonic Mean of the roots, α and β = \(\frac{2\alpha\beta}{\alpha+\beta}\) = \(\frac{2\bigg(\frac{8+2\sqrt5}{5+\sqrt2}\bigg)}{\frac{(4+\sqrt5)}{5+\sqrt2}}\) = \(\frac{4(4+\sqrt5)}{4+\sqrt5}\) = 4.

Let the two numbers be a and b. Then,

A.M = \(\frac{a+b}{2}\)

G.M. = \(\sqrt{ab}\)

H.M. =\(\frac{2ab}{a+b}\)

Given, A.M : G.M = 5 : 4

⇒ \(\frac{\frac{(a+b)}{2}}{\sqrt{ab}}\) = \(\frac{5}{4}\) ⇒ \(\frac{a+b}{2\sqrt{ab}}\) = \(\frac{5}{4}\)                  .....(i)

Also, G.M. – H.M. = \(\frac{16}{5}\)

⇒ \(\sqrt{ab}\) - \(\frac{2ab}{a+b}\) = \(\frac{16}{5}\)                           ......(ii)

From (i), a + b = \(\frac{5\times2}{4}\)\(\sqrt{ab}\) ⇒ a + b \(\frac{5}{2}\)\(\sqrt{ab}\).                      ....(iii)

Putting this value of (a + b) in (ii), we have

\(\sqrt{ab}\) - \(\frac{2ab}{\sqrt{ab}}\) x \(\frac{2}{5}\) = \(\frac{16}{5}\) ⇒ \(\sqrt{ab}\) - \(\frac{4}{5}\)\(\sqrt{ab}\) = \(\frac{16}{5}\)

⇒ \(\frac{1}{5}\)\(\sqrt{ab}\) = \(\frac{16}{5}\) ⇒ \(\sqrt{ab}\) = 16 ⇒ ab = 256

∴ From (iii), a + b = \(\frac{5}{2}\)x 16 = 40.

∴ (a – b)² = (a + b)² – 4ab = 40² – 4 × 256 = 1600 – 1024 = 576

⇒ a – b = ± 24

Now solving a + b = 40 and a – b = ± 24, we get

a = 32, b = 8 or a = 8, b = 32. 

∴ The numbers are 8 and 32.

(d) In H.P.

If a, b, c, d are in A.P. 

⇒ d, c, b, a are in A.P.

⇒ \(\frac{d}{abcd}\), \(\frac{c}{abcd}\), \(\frac{b}{abcd}\), \(\frac{a}{abcd}\) are in A.P. 

(Dividing all terms by abcd)

⇒ \(\frac{1}{abc}\), \(\frac{1}{abd}\), \(\frac{1}{acd}\), \(\frac{1}{bcd}\) are in A.P.

⇒ abc, abd, acd, bcd are in H.P.

a₁, a₂, a₃ ..... , an are in H.P.

⇒ \(\frac{1}{a_1}\),\(\frac{1}{a_2}\),\(\frac{1}{a_3}\) ......., \(\frac{1}{a_n}\) are in A.P

If d is the common difference of the A.P., then

\(\frac{1}{a_2}\) - \(\frac{1}{a_1}\) = d, \(\frac{1}{a_3}\) - \(\frac{1}{a_2}\) = d,......, \(\frac{1}{a_n}\)- \(\frac{1}{a_{n-1}}\) = d

⇒ \(\frac{a_1-a_2}{a_1a_2}\) = d,  \(\frac{a_2-a_3}{a_2a_3}\) = d, ....., \(\frac{a_{n-1}-a_n}{a_{n-1}\,a_n}\) = d

⇒ a₁ – a₂ = a₁a₂d, a₂ – a₃ = a₂a₃d, ..... , a_{n – 1} – a_n = a_{n – 1} a_nd 

⇒ (a₁ – a₂ + a₂ – a₃ + ..... + a_{n – 1} – a_n) = a₁a₂d + a₂a₃d + ..... + a_{n – 1} a_nd

⇒ (a₁ – a_n) = (a₁a₂ + a₂a₃ + ..... + a_{n – 1} a_n)d                          ....(i)

Also, \(\frac{1}{a_1}\),\(\frac{1}{a_2}\),.......\(\frac{1}{a_n}\) is an A.P with common difference d

⇒ \(\frac{1}{a_n}\) = \(\frac{1}{a_1}\) + (n - 1)d ⇒ (n - 1)d = \(\frac{1}{a_n}\) - \(\frac{1}{a_1}\) ⇒ (n - 1)d = \(\frac{a_1-a_n}{a_1a_n}\)

⇒ a₁ – a_n = (n – 1)d a₁a_n                   ...(ii)

From (i) and (ii) 

(n – 1)d a₁a_n = (a₁a₂ + a₂a₃ + .... + a_{n – 1}a_n) d 

⇒ (a₁a₂ + a₂a₃ + .... + a_{n – 1}a_n) = (n – 1) a₁a_n.

(c) 3

Arithmetic mean between a and b is A = \(\frac{a+b}{2}\)

Harmonic mean between a and b is H = \(\frac{2ab}{a+b}\)

Given, A = \(\frac{4}{3}\)H ⇒ \(\frac{a+b}{2}\) = \(\frac{4}{3}\)\(\big(\frac{2ab}{a+b}\big)\)

⇒ 3(a + b)² = 16ab ⇒ 3b² – 10ab + 3a² = 0

⇒ 3\(\big(\frac{b}{a}\big)^2\) - 10\(\big(\frac{b}{a}\big)\) + 3 = 0         (Dividing throughout by a²)

⇒ \(\big(\frac{3b}{a}-1\big)\)\(\big(\frac{b}{a}-3\big)\) = 0

⇒ \(\frac{b}{a}\) = \(\frac{1}{3}\) or \(\frac{b}{a}\) = 3            \(\bigg[\because\,0<a<b\implies\frac{b}{a}\neq\frac{1}{3}\bigg]\)

⇒ \(\frac{b}{a}\) = 3.

Let  \(\frac{a-x}{px}\) = \(\frac{a-y}{qy}\) = \(\frac{a-z}{rz}\) = k

Then, \(\frac{a-x}{px}\) = k ⇒ \(\frac{a-x}{kx}\) = p ⇒ p = \(\frac{1}{k}\)\(\bigg(\frac{a}{x}-1\bigg)\)

Similarly, q = \(\frac{1}{k}\)\(\bigg(\frac{a}{y}-1\bigg)\), r = \(\frac{1}{k}\)\(\bigg(\frac{a}{z}-1\bigg)\)

Now, p, q, r are in A.P

\(\frac{1}{k}\)\(\bigg(\frac{a}{x}-1\bigg)\),\(\frac{1}{k}\)\(\bigg(\frac{a}{y}-1\bigg)\),\(\frac{1}{k}\)\(\bigg(\frac{a}{z}-1\bigg)\) are in A.P.

⇒ \(\frac{a}{x}-1\), \(\frac{a}{y}-1\), \(\frac{a}{z}-1\) are in A.P.                 (Multiplying each term by k)

⇒ \(\frac{a}{x}\), \(\frac{a}{y}\), \(\frac{a}{z}\) are in A.P.                                    (Adding 1 to each term)

⇒ \(\frac{1}{x}\), \(\frac{1}{y}\),\(\frac{1}{z}\) are in A.P.                                    (Dividing each term by a)

⇒ x, y, z are in H.P

x, y, z are in G.P. 

⇒ y² = xz ⇒ 2 log y = log x + log z 

⇒ log x, log y, log z are in A.P. 

⇒ 1 + log x, 1 + log y, 1 + log z are in A.P.    (Adding 1 to each term)

 \(\frac{1}{1+\text{log}\,x}\), \(\frac{1}{1+\text{log}\,y}\), \(\frac{1}{1+\text{log}\,z}\) are in H.p.

a, A₁, A₂, b are in A.P. ⇒ A₁ – a = b – A₂ ⇒ A₁ + A₂ = a + b           ...(i) 

a, G₁, G₂, b are in G.P. ⇒ \(\frac{G_1}{a}\) = \(\frac{b}{G_2}\) ⇒ G₁ G₂ = ab                       ....(ii)

a, H₁, H₂, b are in H.P.

⇒ \(\frac{1}{a}\), \(\frac{1}{H_1}\), \(\frac{1}{H_2}\),\(\frac{1}{b}\) are in A.P.

⇒ \(\frac{1}{H_1}\)- \(\frac{1}{a}\) = \(\frac{1}{b}\) - \(\frac{1}{H_2}\)

⇒ \(\frac{1}{H_1}\) + \(\frac{1}{H_2}\) = \(\frac{1}{a}\) + \(\frac{1}{b}\)

⇒ \(\frac{H_1+H_2}{H_1H_2}\) = \(\frac{a+b}{ab}\) = \(\frac{A_1+A_2}{G_1G_2}\)              (From (i) and (ii))

⇒ \(\frac{G_1G_2}{H_1H_2}\) = \(\frac{A_1+A_2}{H_1+H_2}\). Hence proved.

Now, \(\frac{1}{a}\), \(\frac{1}{H_1}\), \(\frac{1}{H_2}\) are in A.P        (∵ a, H₁, H₂, are in H.P.)

⇒ \(\frac{1}{H_1}\) - \(\frac{1}{a}\) = \(\frac{1}{H_2}\) - \(\frac{1}{H_1}\)

⇒ \(\frac{2}{H_1}\) - \(\frac{1}{H_2}\) = \(\frac{1}{a}\)                             ......(iii)

Also, \(\frac{1}{H_1}\), \(\frac{1}{H_2}\),\(\frac{1}{b}\) are in A.P       (∵ H₁, H₂, b are in H.P.)

⇒ \(\frac{1}{H_2}\) - \(\frac{1}{H_1}\) = \(\frac{1}{b}\) - \(\frac{1}{H_2}\) ⇒ \(\frac{2}{H_2}\) - \(\frac{1}{H_1}\) = \(\frac{1}{b}\)       .....(iv)

Eq. (iii) + 2 x Eqn. (iv) ⇒\(\bigg(\)\(\frac{2}{H_2}\) - \(\frac{1}{H_1}\)\(\bigg)\) +\(\bigg(\)\(\frac{4}{H_2}\) - \(\frac{2}{H_1}\)\(\bigg)\)= \(\frac{1}{a}\) + \(\frac{2}{b}\)

⇒ \(\frac{3}{H_2}\) = \(\frac{b + 2a}{ab}\) ⇒ H₂ = \(\frac{3ab}{2a+b}\)

Eq. (iv) + 2 x Eq. (iii) ⇒ \(\bigg(\)\(\frac{2}{H_2}\) - \(\frac{1}{H_1}\)\(\bigg)\) +\(\bigg(\)\(\frac{4}{H_2}\) - \(\frac{2}{H_1}\)\(\bigg)\)= \(\frac{1}{b}\) + \(\frac{2}{a}\)

⇒ \(\frac{3}{H_1}\) = \(\frac{a + 2b}{ab}\) ⇒ H₁ = \(\frac{3ab}{a+2b}\)

∴ \(\frac{G_1G_2}{H_1H_2}\) = \(\frac{ab}{\frac{3ab}{(a+2b)}\times\frac{3ab}{(2a+b)}}\) = \(\frac{(2a+b)(a+2b)}{9\,ab}\) Hence proved.

(d) \(\frac{11}{2}.\)

Let the terms of the corresponding A.P. be 

a – 2d, a – d, a, a + d, a + 2d. 

Given, a = 1 and \(\frac{a-d}{a+d}\) = \(\frac{1}{2}\)

⇒ 2a – 2d = a + d 

⇒ a = 3d ⇒ d = \(\frac{1}{3}\)a = \(\frac{1}{3}\)           (∵ a = 1)

∴ First three terms of A.P are 1 - \(\frac{2}{3}\), 1 - \(\frac{1}{3}\), 1, i.e., \(\frac{1}{3}\),\(\frac{2}{3}\),1

⇒ First three terms of corresponding H.P are 3, \(\frac{3}{2}\),1

∴ Required sum = 3 + \(\frac{3}{2}\) + 1 = 5\(\frac{1}{2}\) = \(\frac{11}{2}.\)

(b) 4 : 25

Let the two numbers be a and b.

Given, \(\frac{\text{H.M}}{\text{G.M}}\) = \(\frac{20}{29}\) ⇒ \(\frac{\frac{2ab}{a+b}}{\sqrt{ab}}\) = \(\frac{20}{29}\)

⇒ \(\frac{2\sqrt{ab}}{a+b}\) = \(\frac{20}{29}\) ⇒ 58\(\sqrt{ab}\) = 20(a+b)

⇒ 20a - 58\(\sqrt{ab}\) + 20b = 0

⇒ 20\(\frac{a}{b}-58\)\(\sqrt{\frac{a}{b}}\) + 20 = 0                     (Dividing all terms by b)

⇒ 20x² – 58x + 20 = 0 (where x = \(\sqrt{\frac{a}{b}}\))

⇒ 20\(x\)² – 50\(x\) – 8\(x\) + 20 = 0 

⇒ 10\(x\) (2\(x\) – 5) – 4 (2\(x\) – 5) = 0 

⇒ (10\(x\) – 4) (2\(x\) – 5) = 0 

⇒ \(x\) = \(\frac{2}{5}\) or \(\frac{5}{2}\)

⇒ \(\sqrt{\frac{a}{b}}\) = \(\frac{2}{5}\) or \(\frac{5}{2}\)

⇒ \({\frac{a}{b}}\) = \(\frac{4}{25}\) or \(\frac{25}{4}\)

Thus from the given options, the two numbers are in the ratio 4 : 25.

(b) G.P.

Given, 2(y – a) is the H.M. between (y – x) and (y – z) 

⇒ (y – x), 2 (y – a), (y – z) are in H.P

⇒ \(\frac{1}{y-x},\frac{1}{2(y-a)},\frac{1}{y-z}\) are in A.P.

⇒ \(\frac{1}{2(y-a)} - \frac{1}{(y-x)}\) = \(\frac{1}{(y-z)}-\frac{1}{2(y-a)}\)

⇒ \(\frac{y-x-2y+2a}{2(y-a)(y-x)}\) = \(\frac{2y-2a-y+z}{(y-z)\,2(y-a)}\)

⇒ \(\frac{-y-x-2y+2a}{(y-x)}\) = \(\frac{y+z-2a}{(y-z)}\) 

⇒ \(\frac{x+y-2a}{x-y}\) = \(\frac{y+z-2a}{(y-z)}\) 

⇒ \(\frac{(x-a)+(y-a)}{(x-a)-(y-a)}\) = \(\frac{(y-a)+(z-a)}{(y-a)-(z-a)}\)

Now applying componendo and dividendo, we have

\(\bigg(\text{By comp. and div,}\,\frac{x}{y}=\frac{a}{b}\implies\frac{x+y}{x-y}=\frac{a+b}{a-b}\bigg)\)

⇒\(\frac{2(x-a)}{2(y-a)}\) = \(\frac{2(y-a)}{2(z-a)}\) 

⇒ (x – a) (z – a) = (y – a)² 

⇒ (x – a), (y – a), (z – a) are in G.P.

(c) \(\frac{4}{ac}-\frac{3}{b^2}\)

If a, b, c are in H.P, then b = \(\frac{2ac}{a+c}\)

∴ \(\bigg(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\bigg)\)\(\bigg(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\bigg)\)

= \(\bigg(\frac{1}{a}+\frac{a+c}{2ac}-\frac{1}{c}\bigg)\)\(\bigg(\frac{a+c}{2ac}+\frac{1}{c}-\frac{1}{a}\bigg)\)

= \(\bigg(\frac{2c+a+c-2a}{2ac}\bigg)\)\(\bigg(\frac{a+c+2a-2c}{2ac}\bigg)\)

= \(\bigg(\frac{3c-a}{2ac}\bigg)\)\(\bigg(\frac{3a-c}{2ac}\bigg)\) = \(\frac{10ac-3a^2-3c^3}{4a^2c^2}\)

= \(\frac{16ac-(3a^2+3c^2+6ac)}{4a^2c^2}\) = \(\frac{16ac-3(a+b)^2}{4a^2c^2}\)

= \(\frac{4}{ac}-3\bigg(\frac{a+c}{2ac}\bigg)^2\) = \(\frac{4}{ac}-\frac{3}{b^2}\)

(c) Any of these

a, b, c are in A.P. ⇒ 2b = a + c             ...(i) 

a², b², c² are in H.P. ⇒ b² = \(\frac{2a^2c^2}{a^2+c^2}\)           ......(ii)

From eqn (ii) 

b² (a² + c²) = 2a²c² ⇒ b² {(a + c)² – 2ac} = 2a²c² 

⇒ b² {4b² – 2ac} = 2a²c² (From (i) 2b = a + c) 

⇒ 2b⁴ – b2ac – a²c² = 0                 ...(iii) 

⇒ (b² – ac) (2b² + ac) = 0 

⇒ b² – ac = 0 or 2b² + ac = 0

⇒ \(\bigg(\frac{a+c}{2}\bigg)^2\) - ac = 0   b² = \(-\frac{ac}{2}\)

⇒ (a – c)² = 0         a,b, \(-\frac{c}{2}\) are in G.P.

⇒ 2b = 2c (From (i)) 

⇒ b = c.

(a) A.P. 

Let α, β be the roots of the equation ax² + bx + c = 0. 

Then, α + β = \(-\frac{b}{a}\) , αβ = \(\frac{c}{a}\)

Also, given α + β = α² + β²  ⇒ α + β = (α + β)² - 2αβ

⇒ \(-\frac{b}{a}\) = \(\bigg(-\frac{b}{a}\bigg)^2\) - \(\frac{2c}{a}\)

⇒ – ba = b² – 2ac ⇒ b² + ab = 2ac 

⇒ b (b+ a) = 2ac ⇒ \(\frac{b}{c}\) + \(\frac{a}{c}\) = \(\frac{2a}{b}\)

⇒ \(\frac{c}{a},\frac{b}{a},\frac{c}{a}\) are in A.P.

The reciprocals of the terms of the given H.P, i.e., \(\frac{13}{2},6,\frac{11}{2},\)......  form an A.P with first term = \(\frac{13}{2}\) and common difference = 6 - \(\frac{13}{2}\) = \(-\frac{1}{2}.\)

∴ 7th term of this A.P. = a + 6d = \(\frac{13}{2}\) + \(\bigg(-\frac{1}{2}\bigg)\) x 6 = \(\frac{7}{2}.\)

Hence 7th of the given H.P. = \(\frac{1}{\frac{7}{2}}\) = \(\frac{2}{7}.\)