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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

Options

  1. A2
  2. B3
  3. C3 2
  4. D-1

Correct answer

A. 2

Step-by-step solution

Let the progression be a , a+d , a +2 d, Then x₄=3 x₁ a+3 d=3 a 3 d=2 a (i) Again x₇=2 x₃+1 a+6 d=2(a+2 d)+1 2 d=a+1 (ii) Solving (i) and (ii), we get a =3, ~d =2

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