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JEE Advanced 2016 Paper 2 · Mathematics · Sequences and Series · medium

Let a,b,c be three distinct positive real numbers in geometric progression such that a, 2b, 3c are in arithmetic progression.

One or more options correct

Let a,b,ca,b,c be three distinct positive real numbers in geometric progression such that a, 2b, 3ca,\,2b,\,3c are in arithmetic progression. Which of the following statements is/are correct?
  1. A. The common ratio of the geometric progression a,b,ca,b,c is 13\dfrac13
  2. B. The common difference of the arithmetic progression a, 2b, 3ca,\,2b,\,3c is −3c-3c
  3. C. a+b+c=12ca+b+c=12c
  4. D. If the geometric progression a,b,c,…a,b,c,\dots is continued indefinitely, the sum of all its terms is 27c2\dfrac{27c}{2}
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Answer: A, B, D

Write b=arb=ar, c=ar2c=ar^{2} with r>0r>0. The AP condition 2(2b)=a+3c2(2b)=a+3c gives 4ar=a+3ar24ar=a+3ar^{2}, i.e. 3r2−4r+1=03r^{2}-4r+1=0, so (3r−1)(r−1)=0(3r-1)(r-1)=0. Since a,b,ca,b,c are distinct, r≠1r\ne1, hence r=13r=\frac13 and (A) is true. Then a=9ca=9c and b=3cb=3c, so the AP is 9c, 6c, 3c9c,\,6c,\,3c with common difference 6c−9c=−3c6c-9c=-3c; (B) is true. Also a+b+c=9c+3c+c=13c≠12ca+b+c=9c+3c+c=13c\ne12c, so (C) is false. Finally the infinite sum is a1−r=9c2/3=27c2\dfrac{a}{1-r}=\dfrac{9c}{2/3}=\dfrac{27c}{2}, so (D) is true. Hence (A), (B) and (D).
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