Question

If for a sequence, t_n = `(5^("n" - 3))/(2^("n" - 3)`, show that the sequence is a G. P. Find its first term and the common ratio.

Answer 1

The sequence (t_n) is a G.P. if `("t"_("n" + 1))/"t"_"n"` = constant for all n ∈ N.

Now, t_n = `(5^("n" - 3))/(2^("n" - 3)`

∴ t_n+1 = `(5^("n" +1 - 3))/(2^("n" + 1 - 3)) = (5^("n" - 2))/(2^("n" - 2)`

∴ `("t"_("n" + 1))/"t"_"n" = (5^("n" - 2))/(2^("n" - 2)) xx (2^("n" - 3))/(5^("n" - 3)`

= `(5)^(("n" - 2) - ("n" - 3)) xx (2)^(("n" - 3) - ("n" - 2)`

= `(5)^1 xx (2)^(-1)`

= `5/2`

= constant, for all n ∈ N.

∴ The sequence is a G.P. with common ratio (r) = `5/2`

and fiist term = t₁ = `(5^(1 - 3))/(2^(1 - 3)`

= `(5^(-2))/(2^(-2))`

= `2^2/5^2`

= `4/25`