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Question
If for a sequence, tn = `(5^("n" - 3))/(2^("n" - 3)`, show that the sequence is a G. P. Find its first term and the common ratio.
Solution
The sequence (tn) is a G.P. if `("t"_("n" + 1))/"t"_"n"` = constant for all n ∈ N.
Now, tn = `(5^("n" - 3))/(2^("n" - 3)`
∴ tn+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`
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