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Question
For a sequence, if Sn = 2(3n –1), find the nth term, hence show that the sequence is a G.P.
Solution
Sn = 2(3n – 1)
∴ Sn–1 = 2(3n–1 – 1)
But tn = Sn – Sn–1
= 2(3n – 1) – 2(3n–1 – 1)
= 2(3n – 1 – 3n–1 + 1)
= 2(3n – 3n–1)
= 2(3n–1+1 – 3n–1)
∴ tn = 2.3n–1 (3 – 1) = 4.3n–1
∴ tn–1 = `4.3^(("n"– 1) –1)` = 4.3n–2
The sequence (tn) is a G. P.,
If `"t"_"n"/"t"_("n"-1)` = constant
for all n ∈ N
∴ `"t"_"n"/"t"_("n" - 1) = (4.3^("n" - 1))/(4.3^("n" - 2))`
= `3^("n" - 1)/(3^("n" - 1)*3^((-1))`
= 3
= constant for all n ∈ N
∴ r = 3
∴ the sequence is a G.P. with tn = 4.3n–1
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