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प्रश्न
For a sequence, if Sn = 2 (3n – 1), find the nth term, hence show that the sequence is a G.P.
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उत्तर
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(3)n
The sequence (tn) is a G.P. if `("t"_("n" + 1))/"t"_"n"` = constant for all n ∈ N.
∴ `("t"_("n" + 1))/"t"_"n" = (4(3)^"n")/(4(3)^("n" - 1)`
= 3 = constant, for all n ∈ N
∴ the sequence is a G.P. with tn = 4(3)n–1.
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