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QJEEM24Ma43c5: JEE Main 2024 Mathematics - Class 11 JEEM24Ma43c5 Solution: Let three real numbers be in arithmetic progression and be in…

Let three real numbers a,b,ca, b, c be in arithmetic progression and a+1,b,c+3a+1, b, c+3 be in geometric progression. If a>10 and the arithmetic mean of a,ba, b and cc is 8, then the cube of the geometric mean of a,ba, b and cc is

Multiple Choice Options

  • A 120
  • B 316
  • C 312
  • D 128

Step-by-Step Verified Solution

amp;2b=a+c.... (1)amp;b2=(a+1)(c+3).... (2)amp;a+b+c3=8.... (3)\begin{aligned}& 2 b=a+c \quad \text{.... (1)}\\& b^2=(a+1)(c+3) \quad \text{.... (2)}\\& \frac{a+b+c}{3}=8 \quad \text{.... (3)}\end{aligned}

⇒amp;3b3=8amp;b=8⇒amp;ac+3a+c+3=64\begin{aligned}\Rightarrow & \frac{3 b}{3}=8 \\& b=8 \\\Rightarrow \quad & a c+3 a+c+3=64\end{aligned}

amp;3a+c+ac=61... (4)amp;a+c=16amp;c=16−a\begin{aligned}& 3 a+c+a c=61 \quad \text{... (4)}\\& a+c=16 \\& c=16-a\end{aligned}

from equation (4)

amp;3a+16−a+a(16−a)=61amp;⇒(a−15)(a−3)=0amp;a=15(agt;10)amp;⇒a=15,b=8,c=1amp;((a⋅b⋅c)13)3=15×8×1=120\begin{aligned}& 3 a+16-a+a(16-a)=61 \\& \Rightarrow \quad(a-15)(a-3)=0 \\& \quad a=15(a>10) \\& \Rightarrow \quad a=15, b=8, c=1 \\& \left((a \cdot b \cdot c)^{\frac{1}{3}}\right)^3=15 \times 8 \times 1=120\end{aligned}

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