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Question
Answer the following:
For a sequence Sn = 4(7n – 1) verify that the sequence is a G.P.
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Solution
Sn = 4(7n – 1)
∴ Sn–1 = 4(7n–1 – 1)
But, tn = Sn – Sn–1
= 4(7n – 1) – 4(7n–1 – 1)
= 4(7n – 1 – 7n–1 + 1)
= 4(7n – 7n–1)
= 4(7n–1+1 – 7n–1)
= 4.7n–1 (7 – 1)
∴ tn = 24.7n–1
∴ tn–1 = `24.7^(("n" - 1) - 1)`
= 24.7n–2
The sequence is a G.P., if `"t"_"n"/"t"_("n" - 1)` = constant for all n ∈ N.
∴ `"t"_"n"/"t"_("n" - 1) = 24.7^("n" - 1)/24.7^("n" - 2) = 7^("n" - 1)/(7^("n" - 1).7^((-1))`
= 7
= constant, for all n ∈ N
∴ the given sequence is a G.P.
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