What number should be added to 3 5 13 and 19 so that the resulting number may be in proportion?

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What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?

Let the number added be x.

∴ (6 + x) : (15 + x) :: (20 + x) (43 + x)

`=> (6+x)/(15 + x) = (20 + x)/(43 + x)`

⇒ (6 + x)(43 + x) (20 + x)(15 + x)

`=> 258 + 6x + 43x + x^2 = 300 + 20x + 15x + x^2`

`=> 49x - 35x = 300 - 258`

`=> 14x = 42`

`=> x = 3`

Thus, the required number which should be added is 3.

Concept: Concept of Proportion

Is there an error in this question or solution?

(1) 7

(2) 6

(3) –6

(4) –7

Solution:

Let x be the number added to 13, 15 and 19 .

So 13+x, 15+x, 19+x are in HP.

2/(15+x) = 1/(13+x) + 1/(19+x)

2/(15+x) = (13+x+19+x)/(13+x)(19+x)

2/(15+x) = (2x+32)/(13+x)(19+x)

(15+x) (2x+32) = 2(13+x)(19+x)

30x+2x2+480+32x = 2(247+13x+19x+x2)

2x2+62x+480 = 2x2+64x+494

2x = -14

x = -14/2

= -7

Hence option (4) is the answer.

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