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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
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Solution
tn = `(5^("n"-3))/(2^("n"-3)) = (5/2)^("n"-3)`
∴ tn+1 = `(5/2)^("n"+1-3) = (5/2)^("n"-2)`
∴ `("t"_("n"+1))/"t"_"n" = ((5/2)^("n"-2))/((5/2)^("n"-3))`
= `(5/2)^("n" - 2 - "n" + 3)`
= `5/2`, which is a constant
∴ the sequence is a G.P. whose common ratio is `5/2`.
Now, tn = `(5/2)^("n" - 3)`
∴ the first term = t1 = `(5/2)^(1 - 3)`
= `(5/2)^(-2)`
= `(2/5)^2`
= `4/25`
Hence, the first term = t1 = `4/25`
and the common ratio = r = `5/2`.
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