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JEE Main20244 Apr 2024Evening ShiftMathematicsSequences and SeriesActual

Let three real numbers a, b, c be in arithmetic progression and a+1, b, c+3 be in geometric progression. If a>10 and the arithmetic mean of a, b and c is 8, then the cube of the geometric mean of a, b and c is

Options

  1. A128
  2. B316
  3. C120
  4. D312

Correct answer

C. 120

Step-by-step solution

aligned & 2 b=a+c, b^2=(a+1)(c+3) & a+b+c 3 =8 b=8, a+c=16 & 64=(a+1)(19-a)=19+18 a-a^2 & a^2-18 a-45=0 (a-15)(a+3)=0,(a>10) & a=15, c=1, b=8 & ((a b c)^ 1 / 3 )^3=a b c=120 aligned

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