Q84 / 125 · BITSAT 2020 · Mathematics

There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is 6 . If the first and the last numbers are equal then two other numbers are

  • A
    -2,4
  • Correct answer
    B
    -4,2
  • C
    2,6
  • D
    none

Solution

Let the last three numbers in A.P. be a, a +6 , a $+12,$ then the first term is alsoa +12 . But $a+12, a, a+6$ are in G.P. $\therefore \mathrm{a}^{2}=(\mathrm{a}+12)(\mathrm{a}+6) \Rightarrow \mathrm{a}^{2}=\mathrm{a}^{2}+18 \mathrm{a}+72$ $\therefore \mathrm{a}=-4$ $\therefore$ The numbers are 8,-4,2,8.

Hence, the correct answer is option: 'B'.

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