Let three real numbers a, b, c be in arithmetic progression and a+1, b, c+3 be in geometric progression.

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 \(a>10\) and the arithmetic mean of \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) is 8, then the cube of the geometric mean of \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) is

(1) 120

(2) 312

(3) 316

(4) 128

Answer 1

Correct option is (1) 120

\(2 \mathrm{b}=\mathrm{a}+\mathrm{c}, \mathrm{b}^{2}=(\mathrm{a}+1)(\mathrm{c}+3)\)

\(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{3}=8 \rightarrow \mathrm{b}=8, \mathrm{a}+\mathrm{c}=16\)

\(64=(a+1)(19-a)=19+18 a-a^{2}\)

\(a^{2}-18 a-45=0 \rightarrow(a-15)(a+3)=0,(a>10)\)

\(\mathrm{a}=15, \mathrm{c}=1, \mathrm{b}=8\)

\(\left((\mathrm{abc})^{1 / 3}\right)^{3}=\mathrm{abc}=120\)