Question

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Topic - Geometric Progression

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QUESTION QUESTION

⟼ If the sum and product of three numbers in GP are 31 and 125 respectively. ♤ No Spam.

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Answer 1

Answer:

Let the three terms in GP be a/r, a and ar.

Their sum = a/r + a + ar = 31 …(1)

and their product = (a/r)*a*ar = a^3 = 125 …(2)

From (2) a = 5,

5/r + 5 + 5r = 31, or

5/r + 5r = 26, or

5r^2-26r+5=0

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

So r = 1/5 or 5

So the terms are : 1,5 and 25 or 25, 5 and 1.

Answer 2

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➭ Given:-

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

➭ To Find:-

Three number

➭ Soluction :-

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 = ³√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

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

hope it was helpful to you