Question

find three numbers in G.p . such that their sum is 42 and their product is 1728​

Answer 1

Answer:

$\large\bold\red{6,12\: and \:24}$

Step-by-step explanation:

Given,

Three Numbers are in G.P

Let the three numbers be x, y and z .

Now,

They are in GP,

Therefore,

we have,

$xz = {y}^{2}$

Now,

it is Given that,

their product is 1728

Therefore,

we get,

$= > xyz = 1728$

But,

$xz = {y}^{2}$

Therefore,

$= > {y}^{3} = 1728 \\ \\ = > {y}^{3} = {(4 \times 3)}^{3} \\ \\ = > {y}^{3} = {(12)}^{3} \\ \\ = > y = 12$

Also,

it is being given that,

their sum is 42

therefore,

we get,

$= > x + y + z = 42 \\ \\ = > x +12 + z = 42 \\ \\ = > x + z = 42 - 12 \\ \\ = > x + z = 30 \: \: \: \: \: \: ............(i)$

Also,

$xyz = 1728 \\ \\ = > 12x z = 1728 \\ \\ = > xz = \frac{1728}{12} \\ \\ = > xz = 144 \: \: \: \: \: \: \: \: \: ...............(ii)$

From (i) and (ii),

we have,

$x = 6 \\ \\ and \\ \\ z = 24$

Hence,

the required numbers are 6, 12 and 24.

Answer 2

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