Question

(1) If the same number be added to the numbers 2,4,14 and 22 then the resultant
numbers are in proportion. Find the number.


(ii) Find a number such that iſ 25, 7 and 1 are added to it, the product of the first and
third results is equal to the square of the second.

Please answer fast.

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Answer 1

Answers:-

1) Given:

If a number is added to 2 , 4 , 14 , 22 then the resultant numbers are in proportion.

Let the number be x.

So,

The resulting numbers are :

(2 + x) , (4 + x) , (14 + x) , (22 + x).

Now,

These numbers are in proportion.

we know that,

if four numbers/quantities or numbers are in proportion , then product of first quantity & second quantity is equal to the product of second and third quantities.

That is;

Product of extremes = Product of means.

So,

⟹ (2 + x)(22 + x) = (4 + x)(14 + x)

⟹ 2(22 + x) + X(22 + x) = 4(14 + x) + X (14 + x)

⟹ 44 + 2x + 22x + x² = 56 + 4x + 14x + x²

⟹ 24x + 44 = 18x + 56

⟹ 24x - 18x = 56 - 44

⟹ 6x = 12

⟹ x = 12/6

⟹ x = 2

∴ The required number is 2.

_________________________________

2) Let the number which is added be a.

Given:

If a is added to 25 , 7 , 1 ; the product of the first and third results is equal to the square of the second.

i.e.,

(25 + a)(1 + a) = (7 + a)²

• (a + b)² = a² + b² + 2ab

⟹ 25(1 + a) + a(1 + a) = (7)² + (a)² + 14a

⟹ 25 + 25a + a + a² = 49 + a² + 14a

⟹ 26a + 25 = 14a + 49

⟹ 26a - 14a = 49 - 25

⟹ 12a = 24

⟹ a = 24/12

⟹ a = 2

∴ The number that is added is 2.

Answer 2

$\bf \:\large \red{AηsωeR : (i)}$

☆ Let the number added be 'x', so that the numbers

$\sf \: 2+x, \: 4+x, \: 14+x, \: 22+x \: are \: in \: proportion.$

$\tt\implies \:2+x : 4+x \: : : 14+x : 22+x$

$\tt\implies \:\dfrac{2 + x}{4 + x} = \dfrac{14 + x}{22 + x}$

$\tt\implies \:(2 + x)(22 + x) = (4 + x)(14 + x)$

$\tt\implies \:44 + 2x + 22x + {x}^{2} = 56 + 4x + 14x + {x}^{2}$

$\tt\implies \:44 + 24x = 56 + 18x$

$\tt\implies \:24x - 18x = 56 - 44$

$\tt\implies \:6x = 12$

$\tt\implies \:x = 2$

$\tt\implies \: \boxed{ \red{ \bf \:2 \: is \: added \: to \: each \: term. }}$

$\bf \:\large \red{AηsωeR : (ii)}$

☆ Let number added be x, so that numbers become

$\sf \: 25+x, \: 7+x, \: 1+x$

☆ According to the statement, we get

$\tt\implies \: {(7 + x)}^{2} = (25 + x)(1 + x)$

$\tt\implies \:49 + {x}^{2} + 14x = 25 + 25x + x + {x}^{2}$

$\tt\implies \:49 + 14x = 26x + 25$

$\tt\implies \:24x - 12x = 49 - 25$

$\tt\implies \:12x = 24$

$\tt\implies \:x = 2$

$\tt\implies \: \boxed{ \red{ \bf \:2 \: is \: added \: to \: each \: term. }}$