JEE Advanced · Mathematics · 5. Sequences & Series
The minimum value of the sum of real numbers \(a^{-5}, a^{-4}, 3 a^{-3}, 1, a^8\) and \(a^{10}\) with \(a>0\) is
- A 11
- B 34
- C 8
- D 10
Answer & Solution
Correct Answer
(C) 8
Step-by-step Solution
Detailed explanation
Using \(A M \geq G M\),
\(\frac{a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}}{8} \)
\( \geq\left(a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot 1 \cdot a^8 \cdot a^{10}\right)^{\frac{1}{8}} \)
\( \Rightarrow a^{-5}+a^{-4}+3 a^{-3}+1+a^8+a^{10} \geq 8 \cdot 1 \)
Hence, minimum value is 8 .
\(\frac{a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}}{8} \)
\( \geq\left(a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot 1 \cdot a^8 \cdot a^{10}\right)^{\frac{1}{8}} \)
\( \Rightarrow a^{-5}+a^{-4}+3 a^{-3}+1+a^8+a^{10} \geq 8 \cdot 1 \)
Hence, minimum value is 8 .
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