If m arithmetic means (A.M.s) and three geometric means (G.M.s) are inserted between 3 and…

Question

If $m$ arithmetic means (A.M.s) and three geometric means (G.M.s) are inserted between $3$ and $243$ such that the 4th A.M. equals the 2nd G.M., then $m$ is equal to \_\_\_\_\_\_\_\_.

Answer: 39

Accepted Answer

A.P.: $3,A_1,\ldots,A_m,243$ with $d=\dfrac{240}{m+1}$. G.P.: $3,G_1,G_2,G_3,243$ with $r=\left(\dfrac{243}{3}\right)^{1/4}=3$.

$A_4=G_2$: $3+4\left(\dfrac{240}{m+1}\right) = 3(3)^2=27 \implies \dfrac{960}{m+1}=24 \implies m+1=40 \implies m=39.$

Solutions · 1