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Sequences and Series question

2007 · Shift 0 · Q57
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Sequences and Series question

2007 · Shift 0 · Q57

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
In a geometric progression consisting of positive terms, each term equals the sum of the next two terns. Then the common ratio of its progression is equals
  1. A
    5{\sqrt 5 }5​
  2. B
     12(5−1)\,{1 \over 2}\left( {\sqrt 5 - 1} \right)21​(5​−1)
  3. C
    12(1−5){1 \over 2}\left( {1 - \sqrt 5 } \right)21​(1−5​)
  4. D
    125{1 \over 2}\sqrt 521​5​.
View written solutionFree

Correct answer: B

  1. Let the geometric progression be a, ar, ar2, ar3,…a,\, ar,\, ar^2,\, ar^3,\dotsa,ar,ar2,ar3,… where a>0a>0a>0 and since all terms are positive, we must have r>0r>0r>0.

  2. Given condition: each term is equal to the sum of the next two terms.

    So for any term, arn=arn+1+arn+2.ar^n = ar^{n+1} + ar^{n+2}.arn=arn+1+arn+2.

  3. Divide both sides by arnar^narn (nonzero): 1=r+r2.1 = r + r^2.1=r+r2.

  4. Rearrange: r2+r−1=0.r^2 + r - 1 = 0.r2+r−1=0.

  5. Solve the quadratic: r=−1±1+42=−1±52.r = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}.r=2−1±1+4​​=2−1±5​​.

  6. Since all terms of the GP are positive, the common ratio must be positive. Hence, r=5−12.r = \frac{\sqrt{5}-1}{2}.r=25​−1​.

  7. Compare with options:

    • A: 5\sqrt{5}5​ ❌
    • B: 12(5−1)\dfrac{1}{2}(\sqrt{5}-1)21​(5​−1) ✅
    • C: 12(1−5)\dfrac{1}{2}(1-\sqrt{5})21​(1−5​) ❌ negative
    • D: 52\dfrac{\sqrt{5}}{2}25​​ ❌

Therefore, the correct option is B.

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