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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2016 · Mathematics · Sequences and Series

The fourth term of an A.P. is three times of the first term and the...

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. C32\frac{3}{2}
  4. D-1

Correct answer

A. 2

Step-by-step solution

Let the progression be aa, a+da+d, a +2 d, Then x4=3x1⇒a+3d=3a⇒3d=2a…x_{4}=3 x_{1} \Rightarrow a+3 d=3 a \Rightarrow 3 d=2 a \ldots (i) Again x7=2x3+1⇒a+6d=2(a+2d)+1⇒2d=a+1…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 a=3,d=2\mathrm{a}=3, \mathrm{ d}=2
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