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Answer the following: For a sequence Sn = 4(7n – 1) verify that the sequence is a G.P. - Mathematics and Statistics

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प्रश्न

Answer the following:

For a sequence Sn = 4(7n – 1) verify that the sequence is a G.P.

योग
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उत्तर

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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अध्याय 2: Sequences and Series - Miscellaneous Exercise 2.2 [पृष्ठ ४१]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 2 Sequences and Series
Miscellaneous Exercise 2.2 | Q II. (7) | पृष्ठ ४१

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