Question

Suppose 2,7,9, are subtracted respectively from first, second
third and fourth term of a Geometric progression consisting of
four numbers, the resulting numbers are found to be in AP,
then the smallest of the one numbers in the G.P is
​

Answer 1

The smallest number is (- 24).

Complete question:

Suppose 2, 7, 9, 5 are subtracted respectively from first, second, third and fourth term of a Geometric progression consisting of four numbers, the resulting numbers are found to be in AP, then the smallest one of these numbers in the G.P is = ?

Step-by-step explanation:

Let the four numbers in GP be

a, ar, ar², ar³

Given that, 2, 7, 9, 5 are subtracted respectively from the above four numbers. Then the four numbers become

a - 2, ar - 7, ar² - 9, ar³ - 5

Here a - 2, ar - 7, ar² - 9, ar³ - 5 are in AP. Then

2 (ar - 7) = a - 2 + ar² - 9

or, 2ar - 14 = a + ar² - 11

or, a (1 - 2r + r²) = - 3 ..... (1)

2 (ar² - 9) = ar - 7 + ar² - 5

or, 2ar² - 18 = ar + ar³ - 12

or, ar (1 - 2r + r²) = - 6

or, r (- 3) = - 6 [ by (1) ]

or, - 3r = - 6

or, r = 2

Putting r = 2 in (1), we get

a (1 - 4 + 4) = - 3

or, a = - 3

Thus the given GP is

- 3, (- 3) * 2, (- 3) * 2², (- 3) * 2³

i.e., -3, - 6, - 12, - 24

∴ the smallest of these numbers is (- 24).

Related problems:

• 1. If the first three terms of an AP are b, c and 2b, find the ratio of b and c. - https://brainly.in/question/12585983
• 2. Which term of the GP 2,1,1/2,1/4,.... Is 1/1024? - https://brainly.in/question/2354435

Answer 2

the smallest of the one numbers in the G.P is -24

Step-by-step explanation:

Let the four numbers in GP be

a, ar, ar², ar³      (say)

According to the question,

2, 7, 9, 5 are subtracted respectively from the above four numbers.

Then the four numbers become a - 2, ar - 7, ar² - 9, ar³ - 5

Now by the question we get,

a - 2, ar - 7, ar² - 9, ar³ - 5 are in AP.

As they are in AP so the common difference is equal. Therefore we get,

2 (ar - 7) = a - 2 + ar² - 9

⇒ 2ar - 14 = a + ar² - 11

⇒ a (1 - 2r + r²) = - 3 ..... (1)

2 (ar² - 9) = ar - 7 + ar² - 5

⇒ 2ar² - 18 = ar + ar³ - 12

⇒ ar (1 - 2r + r²) = - 6

⇒ r (- 3) = - 6 [ by equation(1) ]

⇒ - 3r = - 6

⇒ r = 2

Putting r = 2 in (1), we get

a (1 - 4 + 4) = - 3

⇒ a = - 3

Thus the given GP is

- 3, (- 3) $\times$ 2, (- 3) $\times$ 2², (- 3) $\times$ 2³

⇒ -3, - 6, - 12, - 24

∴ the smallest of these numbers is  -24