# the n term of an arithmetic progression is Tn . Show that Un=5/2(-2)^2 [(10-Tn)/17] is the n term of a geometric progression.

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You need to write the equation that relates three consecutive terms of arithmetic progression such that:

`T_n = (T_(n-1) + T_(n+1))/2 =gt 2T_n = T_(n-1) + T_(n+1)`

You also need to write the equation that relates three consecutive terms of geometric progression such that:

`U_n = sqrt(U_(n-1)*U_(n+1)) =gt (U_n)^2 =U_(n-1)*U_(n+1)`

Hence, you need to check that `(5/2(-2)^2 ((10-Tn)/17))^2 = 5/2(-2)^2 ((10-T(n-1))/17)*5/2(-2)^2 ((10-T(n+1))/17)` `(25/4*16)*(10-Tn)^2/17^2 = 25/64(10-T(n-1))/17*(10-T(n+1))/17`

`(10-Tn)^2/17^2 = ((10-T(n-1))*(10-T(n+1)))/17^2`

`(10-Tn)^2 = (10-T(n-1))*(10-T(n+1))`

Expanding the square and opening the brackets yields:

`10^2 - 20T_n + (T_n)^2 = 10^2 - 10*T_(n+1) - 10*T_(n-1) + (T(n-1)*T(n+1))`

Reducing like terms yields:

`(T_n)^2 - 20T_n = (T(n-1)*T(n+1)) - 10(T_(n+1) + T_(n-1))`

You need to substitute `2T_n` for `T_(n+1) + T_(n-1)` such that:

`(T_n)^2 - 20T_n = (T(n-1)*T(n+1)) - 10*(2T_n)`

`(T_n)^2 - 20T_n = (T(n-1)*T(n+1)) - 20T_n `

Reducing like terms yields:

`(T_n)^2 = (T(n-1)*T(n+1)) `

**Notice that the last equation proves that the term `U_n` gives the general term of a geometric progression, hence, `U_n =5/2(-2)^2 ((10-Tn)/17)` represents the n-th term of a geometric progression.**