Let x, x, x, x be in a geometric progression. 2, 7, 9, 5 are subtracted respectively from x, x, x x then the 1 2 3 4 1 2 3 4 resulting numbers are in an arithmetic progression. Then the value of 1 24$($$x_{1}$$$x_{2}$$$x_{3}$$$$$$x_{4}$$)$is:
Step-by-Step Solution
Key Concept: Let the original GP be$a,ar,ar^2,ar^3$and apply the subtractions to form the new condition.
$x_{1}$,$x_{2}$,$x_{3}$,$x_{4}$$\to G.P.$(4)$Let a, ar, ar, ar$$\to G.P. 2 3 Now$a - 2$,$ar - 7$,$a - 9$, ar - 5$$\to A.P. 2 3 2($ar - 7) = a - 2 + ar$$2 - 9$....(i) 2 (ar$2 -9) = ar - 7 + ar$$3 - 5$....(ii) Solving$r = 2$, a = -3 4 6$∴$Product = x_{1}$,$x_{2}$,$x_{3}$,$x_{4} = a$$r = 81$$\times$ 64 Solving$r = 2$, a = -3 4 6 ∴$Product = x_{1}$,$x_{2}$,$x_{3}$,$x_{4} = a$$r = 81$$\times$ 64
Correct Answer: 4