Sequence and Series

Concept:Use the properties of a geometric progression (GP) and the given sum and arithmetic mean to find the common ratio and first term.Explanation:Let the GP be $a$, $ar$, $ar^2$, $ar^3$.Sum: $a(1+r+r^2+r^3)=60$.AM of first and last: $\frac{a+ar^3}{2}=18 \Rightarrow a(1+r^3)=36$.Factor: $1+r^3=(1+r)(1-r+r^2)$ and $1+r+r^2+r^3=(1+r)(1+r^2)$.Divide the sum equation by the AM equation: $\frac{(1+r)(1+r^2)}{(1+r)(1-r+r^2)} = \frac{60}{36} = \frac5 3$.Simplify: $\frac{1+r^2}{1-r+r^2} = \frac5 3 \Rightarrow 3(1+r^2)=5(1-r+r^2) \Rightarrow 2r^2-5r+2=0$.Solve: $(2r-1)(r-2)=0 \Rightarrow r=2$ or $r=\frac12$.If $r=2$: $a(1+8)=9a=36 \Rightarrow a=4$. Numbers: $4, 8, 16, 32$. Sum = 60, AM = 18. Verified.If $r=\frac12$: $a(1+\frac18)=\frac98a=36 \Rightarrow a=32$. Numbers: $32, 16, 8, 4$ (decreasing order; not listed as such).Thus the required numbers are $4, 8, 16, 32$, which matches option A.Answer:Option A: $4, 8, 16, 32$