JEE Mains · Maths · STD 11 - 8. sequence and series
Let three real numbers \(a, b, c\) be in arithmetic progression and \(\mathrm{a}+1, \mathrm{~b}, \mathrm{c}+3\) be in geometric progression. If \(\mathrm{a}>10\) and the arithmetic mean of \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) is \(8\) , then the cube of the geometric mean of \(a, b\) and \(c\) is
- A \(120\)
- B \(312\)
- C \(316\)
- D \(128\)
Answer & Solution
Correct Answer
(A) \(120\)
Step-by-step Solution
Detailed explanation
\( 2 b=a+c, b^2=(a+1)(c+3) \) \( \frac{a+b+c}{3}=8 \rightarrow b=8, a+c=16 \) \( 64=(a+1)(19-a)=19+18 a-a^2 \) \( a^2-18 a-45=0 \rightarrow(a-15)(a+3)=0,(a>10) \) \( a=15, c=1, b=8 \) \( \left((a b c)^{1 / 3}\right)^3=a b c=120\)
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