The nth term of an arithmetic progression is T_n , Show that $\displaystyle U_n=\frac{5}{2}(-2)^2(\frac{10-T_n}{17} )$ is the nth term of a geometric progression .

Printable View

- Sep 8th 2009, 03:58 AMthereddevilsProgression
The nth term of an arithmetic progression is T_n , Show that $\displaystyle U_n=\frac{5}{2}(-2)^2(\frac{10-T_n}{17} )$ is the nth term of a geometric progression .