Q50 / 125 · BITSAT 2016 · Mathematics

The fourth term of an A.P. is three times of the first term and the seventh term exceeds the twice of the third term by one, then the common difference of the progression is

  • Correct answer
    A
    2
  • B
    3
  • C
    $\frac{3}{2}$
  • D
    -1

Solution

Let the progression be $a$, $a+d$, a +2 d, Then $x_{4}=3 x_{1} \Rightarrow a+3 d=3 a \Rightarrow 3 d=2 a \ldots$ (i) Again $x_{7}=2 x_{3}+1$ $\Rightarrow a+6 d=2(a+2 d)+1 \Rightarrow 2 d=a+1 \quad \ldots$ (ii) Solving (i) and (ii), we get $\mathrm{a}=3, \mathrm{~d}=2$

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

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