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 of the four numbers in the G.P TS.​

Answer 1

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

Given:

• 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.

To find:

• Smallest of the four numbers in GP

Answer:

• Let the four numbers in GP be a, ar, ar² and ar³.
• After subtracting, we get: a - 2, ar - 7, ar² - 9, ar³ - 5
• As a - 2, ar - 7, ar² - 9, ar³ - 5 are in AP,

              2 (ar - 7) = (a - 2) + (ar² - 9)

          ⇒ 2ar - 14 = a + ar² - 11

          ⇒ a (1 - 2r + r²) = - 3   ------- Equation 1

              2 (ar² - 9) = (ar - 7) + (ar² - 5)

          ⇒ 2ar² - 18 = ar + ar³ - 12

          ⇒ ar (1 - 2r + r²) = - 6    ------------ Equation 2

• Dividing Equation 1 by Equation 2, we get: r = 2
• Putting the value of r in Equation 1, we get: a = -3

• Numbers in GP = -3 , (-3)*2, (-3) * 4, (-3) *8 = -3, -6, -12, -24

• Smallest of the four numbers in GP = -24