Question

#Quality Question
@Sequence And Series

The sum and product of three numbers in G.P. are 31 and 125 respectively.

Find The Three Numbers

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Answer 1

▪ Given :-

Three Numbers are in G.P.

such that Their

• Sum = 31

• Product = 125

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▪ To Find :-

• The Three Numbers

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▪ Solution :-

Let the Three Numbers in G.P.  be :

$\bf \dfrac{a}{r} \:  , \:  a \: and\: ar$

Where r is common ratio of corresponding G.P.

According to the Given Condition :

$\bf  \dfrac{a}{r}  + a + ar = 31 \:  \:  \:  \: . \:  .\:  .\:  \{i \}$

Also ,

$\sf \dfrac{a}{r}.a.ar = 125 \\  \\  :\longmapsto \sf  {a}^{3}  = 125 \\  \\  \sf :\longmapsto a =  \sqrt[3]{125}  \\  \\  \Large \purple{ :\longmapsto  \underline {\boxed{{\bf a = 5} }}}$

Putting Value of a in {i} We get  ,

$\sf\dfrac{5}{r}  + 5 + 5r = 31 \\  \\  :\longmapsto \sf\dfrac{5 + 5r + 5 {r}^{2} }{r}  = 31 \\  \\ :\longmapsto \sf5 {r}^{2}  + 5r + 5 = 31r \\  \\ :\longmapsto \large \bf5 {r}^{2}  - 26r + 5 = 0 \\  \\:\longmapsto  \sf {5r}^{2}  - 25r - r + 5 = 0 \\  \\ :\longmapsto \sf5r(r - 5) - 1(r - 5) = 0 \\  \\ :\longmapsto \sf(r - 5)(5r - 1) \\  \\  \Large\bf \purple{ :\longmapsto \underline {\boxed{{\bf r = 5 \:  \:  \: or \:  \:  \: r =  \frac{1}{5} } }}}$

Case - 1 》

If r = 5

Numbers will be :

1 , 5 and 25

Case - 2》

If r $=  \bf\dfrac{1}{5}$

Numbers will be :

25 , 5 and 1

Combining Both Cases We Get,

The three numbers are

$\huge \mathfrak{1,5 \: and \: 25}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Given :-

The sum and product of three numbers in G.P. are 31 and 125 respectively.

To Find :-

Three number

Solution :-

Let us assume that the three numbers are a, ar and a/r

a + ar + a/r = 31 (1)

a × ar × a/r = 125

a × a × a = 125

a³ = 125

a = ∛125

a = 5

Now, Put the value of a in the first equation, We get

5 + 5r + 5/r = 31

5r² + 5r + 5/r = 31

5r² + 5r + 5 = 31(r)

5r² + 5r + 5 - 31r = 0

5r² - 26r + 5 = 0

5r² - (25r + r) + 5 = 0

5r² - 25r - r + 5 = 0

5r(r - 5) - 1(r - 5) = 0

(r - 5)(5r - 1) = 0

So,

Either

r - 5 = 0

r = 0 + 5

r = 5

Or,

5r - 1 = 0

5r = 0 + 1

5r = 1

r = 1/5

Putting r = 5

a = 5

ar = 5(5) = 25

a/r = 5/5 = 1

Putting r = 1/5

a = 5

ar = 5(1/5) = 1

ar = 5/(1/5) = 5/1 × 5/1 = 25/1 = 25

Therefore

numbers are = 25, 1, 5