Question

The product of first three terms of a G.P is 1000.If 6 is added to its second term and 7 added to its third term ,the terms become in A.P .Find the G.P.

Answer 1

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GiveN:

• The product of first three terms of a G.P is 1000.
• 1000.If 6 is added to its second term and 7 added to its third term, the terms form in AP.

To FinD:

• The GP?

Step-by-step Explanation:

We have the three terms in GP.

Let the terms be a/r, a, ar

According to question,

⇒ a/r × a × ar = 1000

⇒ a³ = 1000

⇒ a³ = 10³

⇒ a = 10

Now it is given that when 6 is added to its second term and 7 added to its third term, the terms form in AP.

• The second term will be a + 6
• And third term will be ar + 7

And we know that when the terms form in AP,

• 2(Second term) = First term + Third term

Plugging the given values:

⇒ 2(a + 6) = a/r + ar + 7

⇒ 2(10 + 6) = 10/r + 10r + 7

⇒ 32 = 10/r + 10r + 7

⇒ 10/r + 10r = 25

⇒ 10 + 10r² / r = 25

⇒ 10r² + 10 = 25r

⇒ 2r² - 5r + 2 = 0

Finding values of r by middle term factorisation,

⇒ 2r² - 4r - r + 2 = 0

⇒ 2r(r - 2) - 1(r - 2) = 0

⇒ (2r - 1)(r - 2) = 0

⇒ Then, r = 1/2 , 2

Hence,

• The numbers are 5,10,20 / 20,10,5.

Answer 2

Step-by-step explanation:

Given :

• The product of first three terms of a G.P is 1000.

• If 6 is added to its second term and 7 added to its third term

To Find :

• the terms become in A.P .Find the G.P.

Solution :

Let the three numbers = a, ar and a/r

• According to the Question :

The product of first three terms of a G.P is 1000.

• Substitute all values :

➭ a × ar × a/r = 1000

➭ a × a × a = 1000

➭ a³ =1000

➭ a = 10

Thus, 2b = a+ c are in AP

Let a/ r, a + 6, ar +7 will form an AP

= 2(a + 6) = a/ r + ( ar + 7)

• Substitute all values :

➭ 2 (10 + 6) = 10/r + (10r + 7)

➭ 2(16) = 10/r + 10r +7

➭ 32 = 10/r + 10r + 7

➭ 32 - 7 = 10/r + 10r

➭ 25 = 10/ r + 10r

➭ 5 = 2/r + 2r

➭ 5r = 2 + 2r²

Factorizing the equation :

r = 1/2 or r = 2

• Thus, if r = 1/2, then GP is 20, 10,5 If r = 2,  then GP is 5,10,20.