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EXERCISE 2.1 · Q12

Q.If for a sequence, tn=5n−32n−3t_n=\dfrac{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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tn+1tn=5(n+1)−3/2(n+1)−35n−3/2n−3=5n−22n−2×2n−35n−3=52\dfrac{t_{n+1}}{t_n}=\dfrac{5^{(n+1)-3}/2^{(n+1)-3}}{5^{n-3}/2^{n-3}}=\dfrac{5^{n-2}}{2^{n-2}}\times\dfrac{2^{n-3}}{5^{n-3}}=\dfrac52, a constant independent of nn, so the sequence is a G.P. with r=52r=\dfrac52. The first term is $t_1=\dfrac{5^{-2} …

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