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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
Question
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
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Answer

a
$ 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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