Solution:

Let that number be x.

∴ \(\displaystyle \frac{12 + x}{16 + x} = \frac {16 + x}{21 + x}\)

∴ (12 + x)(21 + x) = (16 + x)(16 + x)

∴ 12(21 + x) + x(21 + x) = 16(16 + x) + x(16 + x)

∴ 252 + 12x + 21x + x² = 256 + 16x + 16x + x²

∴ 252 + 33x = 256 + 32x

∴ 33x − 32x = 256 − 252

∴ x = 4

∴ 4 should be added to 12, 16 and 21 so that resultant numbers are in continued proportion.

Solution:

(23 − x) is the mean proportional of (28 − x) and (19 − x).

∴ \(\displaystyle \frac{28 - x}{23 - x} = \frac {23 - x}{19 - x}\)

∴ (28 − x)(19 − x) = (23 − x)(23 − x)

∴ 28(19 − x) − x(19 − x) = 23(23 − x) − x(23 − x)

∴ 532 − 28x − 19x + x² = 529 − 23x − 23x + x²

∴ 532 − 47x = 529 − 46x

∴ − 47x + 46x = 529 − 532

∴ − x = − 3

i.e. x = 3

Solution:

Let, the first number be x.

The second number is 12. ... (Given)

The third number = 26 − x.

Now, \(\displaystyle \frac{x}{12} = \frac {12}{26 - x}\)

∴ x(26 − x) = 12 × 12

∴ 26x − x² = 144

∴ 0 = 144 − 26x + x²

i.e. x² − 26x + 144 = 0

∴ x² − 18x − 8x + 144 = 0

∴ x(x − 18) − 8(x − 18) = 0

∴ (x − 18)(x − 8) = 0

∴ x − 18 = 0 OR x − 8 = 0

∴ x = 18 OR x = 8

∴ The first number is 18 OR 8.

∴ The third number
= 26 − x = 26 − 18 = 8
OR
The third number
= 26 − x = 26 − 8 = 18.

∴ Those numbers are 18, 12, 8 or 8, 12, 18.

Solution:

(a + b + c)(a − b + c) = a² +  b² + c²

∴ a² − ab + ac + ab − b² + bc + ac − bc + c² =  a² +  b² + c²

∴ 2ac − b² = b²

∴ 2ac = b² + b²

∴ 2ac = 2b²

∴ ac = b²

∴ a × c = b × b

∴ \(\displaystyle \frac{a}{b} = \frac {b}{c}\)

∴ a, b, c are in continued proportion.

Solution:

\(\displaystyle \frac{a}{b} = \frac {b}{c} = k \text{ say}\)

∴ b = ck ... (I)

and a = bk
∴ a = ck × k
∴ a = ck² ... (II)

(i)

(a + b + c) (b − c) = ab − c²

Now, LHS
= (a + b + c) (b − c)
= (ck² + ck + c) (ck − c)
= c(k² + k + 1) × c(k − 1)
= c²(k³ − k² + k² − k + k − 1)
= c²(k³ − k² + k² − k + k − 1)
= c²(k³ − 1) ... (iii)

And RHS
= ck² × ck − c²
= c²k³ − c²
= c²(k³ − 1) ... (iv)

From (iii) and (iv),
 LHS = RHS
∴ (a + b + c) (b − c) = ab − c²

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(ii)

(a² + b²)(b² + c²) = (ab + bc)²

Now, LHS
= (a² + b²)(b² + c²)
= [(ck²)² + (ck)²][(ck)² + c²]
= (c²k⁴ + c²k²)(c²k² + c²)
= c²k²(k² + 1) × c²(k² + 1)
= c⁴k²(k² + 1)² ... (v)

And RHS
= (ab + bc)²
= (ck² × ck + ck × c)²
= (c²k³ + c²k)²
= c⁴k²(k² + 1)² ... (vi)

From (v) and (vi),
 LHS = RHS
∴ (a² + b²)(b² + c²) = (ab + bc)²

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(iii)

\(\displaystyle \frac{a^2 + b^2}{ab} = \frac {a + c}{b}\)

Now, LHS

= \(\displaystyle \frac{a^2 + b^2}{ab}\)

= \(\displaystyle \frac{c^2k^4 + c^2k^2}{c^2k^3}\)

= \(\displaystyle \frac{c^2k^2(k^2 + 1)}{c^2k^3}\)

= \(\displaystyle \frac{k^2 + 1}{k}\) ... (vii)

And RHS

= \(\displaystyle \frac {a + c}{b}\)

= \(\displaystyle \frac{ck^2 + c}{ck}\)

= \(\displaystyle \frac{c(k^2 + 1)}{ck}\)

= \(\displaystyle \frac{k^2 + 1}{k}\) ... (viii)

From (vii) and (viii),
 LHS = RHS

∴ \(\displaystyle \frac{a^2 + b^2}{ab} = \frac {a + c}{b}\)

Solution:

Let, the mean proportional be m.

∴ \(\displaystyle \frac {\displaystyle\frac{x + y}{x - y}}{m}\) = \(\displaystyle \frac {m}{\displaystyle\frac{x^2 - y^2}{x^2y^2}}\)

∴ \(\displaystyle \frac{x + y}{x - y} \times \frac{x^2 - y^2}{x^2y^2} = m \times m\)

∴ \(\displaystyle \frac{x + y}{x - y} \times \frac{(x + y)(x - y)}{x^2y^2} = m^2\)

∴ \(\displaystyle \frac{(x + y)^2}{x^2y^2} = m^2\)

Taking square roots of both sides,

∴ \(\displaystyle \frac{x + y}{xy} = m\)

i.e. \(\displaystyle m = \frac{x + y}{xy}\)

∴ The mean proportional is \(\displaystyle \frac{x + y}{xy}\).