Let x1, x2, x3, x4 be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from x1, x2, x3, x4, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1 x2 x3 x4) is:
- 18
- 216
- 36
- 72
Answer
(2) 216
Solution
Let x1 = a, x2 = ar, x3 = ar2, x4 = ar3.
a - 2, ar - 7, ar2 - 9, ar3 - 5 are in A.P.
From the first three terms: 2(ar - 7) = (a - 2) + (ar2 - 9)
a(r2 - 2r + 1) = -3 a(r-1)2 = -3 ... (i)
From the last three terms: 2(ar2 - 9) = (ar - 7) + (ar3 - 5)
ar(r2 - 2r + 1) = -6 ar(r-1)2 = -6 ... (ii)
(ii) ÷ (i): r = 2, then a = -3
G.P.: -3, -6, -12, -24 (check: -5, -13, -21, -29 is an A.P. ✓)
x1 x2 x3 x4 = (-3)(-6)(-12)(-24) = 5184
124(x1 x2 x3 x4) = 518424 = 216
Answer: (2)