SATHEE: Sequences and Series 3 Question 9

Sequences and Series 3 Question 9

9. Let $a, b, c$ be in an AP and $a^{2}, b^{2}, c^{2}$ be in GP. If $a < b < c$ and $a + b + c = \frac{3}{2}$ , then the value of $a$ is

(2002, 1M)

(a) $\frac{1}{2 \sqrt{2}}$

(b) $\frac{1}{2 \sqrt{3}}$

(d) $\frac{1}{2} - \frac{1}{\sqrt{2}}$

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Answer:

Correct Answer: 9. (a)

Solution:

Since, $a, b$ and $c$ are in an AP.

Let $a = A - D, b = A, c = A + D$

Given, $a + b + c = \frac{3}{2}$

$\Rightarrow (A - D) + A + (A + D) = \frac{3}{2}$

$\Rightarrow 3 A = \frac{3}{2} \Rightarrow A = \frac{1}{2}$

$∴$ The number are $\frac{1}{2} - D, \frac{1}{2}, \frac{1}{2} + D$ .

Also, $\frac{1}{2} - D^{2}, \frac{1}{4}, \frac{1}{2} + D^{2}$ are in GP.

$∴ \frac{1}{4}^{2} = \frac{1}{2} - D^{2} \frac{1}{2} + D^{2} \Rightarrow \frac{1}{16} = \frac{1}{4} - D^{2}$

$\Rightarrow \frac{1}{4} - D^{2} = \pm \frac{1}{4} \Rightarrow D^{2} = \frac{1}{2} \Rightarrow D = \pm \frac{1}{\sqrt{2}}$

$∴ a = \frac{1}{2} \pm \frac{1}{\sqrt{2}}$

So, out of the given values, $a = \frac{1}{2} - \frac{1}{\sqrt{2}}$ is the right choice.

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Sequences and Series 3 Question 9

9. Let $a, b, c$ be in an AP and $a^{2}, b^{2}, c^{2}$ be in GP. If $a < b < c$ and $a + b + c = \frac{3}{2}$ , then the value of $a$ is

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Sequences and Series 3 Question 9

9. Let a,b,c be in an AP and a2,b2,c2 be in GP. If a<b<c and a+b+c=32, then the value of a is

Answer:

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चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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9. Let $a, b, c$ be in an AP and $a^{2}, b^{2}, c^{2}$ be in GP. If $a < b < c$ and $a + b + c = \frac{3}{2}$ , then the value of $a$ is