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Sequences and Series question

2014 · Shift 0 · Q39
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Sequences and Series question

2014 · Shift 0 · Q39

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :
  1. A
    2−32 - \sqrt 32−3​
  2. B
    2+32 + \sqrt 32+3​
  3. C
    2+3\sqrt 2 + \sqrt 32​+3​
  4. D
    3+23 + \sqrt 23+2​
View written solutionFree

Correct answer: B

  1. Let the three numbers in increasing G.P. be ar, a, ar\frac{a}{r},\ a,\ arra​, a, ar where a>0a>0a>0 and since the numbers are increasing and positive, we must have r>1.r>1.r>1.

  2. After doubling the middle term, the new numbers become ar, 2a, ar.\frac{a}{r},\ 2a,\ ar.ra​, 2a, ar. These are given to be in A.P.

  3. Condition for A.P. In an arithmetic progression, the middle term is the average of the other two terms: 2(2a)=ar+ar.2(2a)=\frac{a}{r}+ar.2(2a)=ra​+ar. So, 4a=ar+ar.4a=\frac{a}{r}+ar.4a=ra​+ar.

  4. Simplify Divide throughout by aaa (a>0)(a>0)(a>0): 4=1r+r.4=\frac{1}{r}+r.4=r1​+r. Multiply by rrr: 4r=1+r2.4r=1+r^2.4r=1+r2. Rearranging, r2−4r+1=0.r^2-4r+1=0.r2−4r+1=0.

  5. Solve the quadratic r=4±16−42=4±122=4±232=2±3.r=\frac{4\pm\sqrt{16-4}}{2}=\frac{4\pm\sqrt{12}}{2}=\frac{4\pm 2\sqrt3}{2}=2\pm\sqrt3.r=24±16−4​​=24±12​​=24±23​​=2±3​.

  6. Use the increasing G.P. condition Since the G.P. is increasing and positive, r>1r>1r>1. Now, 2−3<1,2-\sqrt3<1,2−3​<1, so it is not possible. Hence, r=2+3.r=2+\sqrt3.r=2+3​.

  7. Check options

    • A: 2−32-\sqrt32−3​ ✗
    • B: 2+32+\sqrt32+3​ ✓
    • C: 2+3\sqrt2+\sqrt32​+3​ ✗
    • D: 3+23+\sqrt23+2​ ✗

Therefore, the common ratio is 2+3.\boxed{2+\sqrt3}.2+3​​.

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