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JEE Mains · Maths · STD 11 - 8. sequence and series
If \(a, b, c\) are in \(A.P.\) and \(a^2, b^2, c^2\) are in \(G.P.\) such that \( a < b\) \( < c\) and \(a+b+c\,= \frac{3}{4}\) , then the value of \(a\) is
- A \(\frac{1}{4} - \frac{1}{{3\sqrt 2 }}\)
- B \(\frac{1}{4} - \frac{1}{{4\sqrt 2 }}\)
- C \(\frac{1}{4} - \frac{1}{{\sqrt 2 }}\)
- D \(\frac{1}{4} - \frac{1}{{2\sqrt 2 }}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4} - \frac{1}{{2\sqrt 2 }}\)
Step-by-step Solution
Detailed explanation
\(\because a,b,c\) are in \(A.P.\) then \(a+c=2b\) also it is given that, \(a + b + c = \frac{3}{4}\,\,\,\,\,......\left( 1 \right)\) \( \Rightarrow 2b + b = \frac{3}{4}\,\,\,\, \Rightarrow b = \frac{1}{4}\,\,\,\,\,\,\,....\left( 2 \right)\) Again it is given that,…
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