Q51 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics

There are four numbers of which the first three are in GP and the last three are in AP, 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 of the above

Solution

Let 3 numbers in AP are $a,(a+6),(a+12)$ Also, first and last number out of 4 numbers are equal $\therefore 4$ numbers are $(a+12), a,(a+6),(a+12)$ Also, given, first 3 numbers are in GP $\begin{array}{ll} \Rightarrow & a^2=(a+12)(a+6) \\ \Rightarrow & a^2=a^2+18 a+72 \\ \Rightarrow & a=-\frac{72}{18}=-4 \end{array}$ $\therefore 4$ numbers are $8,-4,2,8$

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

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