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If for a sequence, tn = 5n-32n-3, show that the sequence is a G.P. Find its first term and the common ratio

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प्रश्न

If for a sequence, tn = `(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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उत्तर

tn = `(5^("n"-3))/(2^("n"-3)) = (5/2)^("n"-3)`

∴ tn+1 = `(5/2)^("n"+1-3) = (5/2)^("n"-2)`

∴ `("t"_("n"+1))/"t"_"n" = ((5/2)^("n"-2))/((5/2)^("n"-3))`

= `(5/2)^("n" - 2 - "n" + 3)`

= `5/2`, which is a constant

∴ the sequence is a G.P. whose common ratio is `5/2`.

Now, tn = `(5/2)^("n" - 3)`

∴ the first term = t1 = `(5/2)^(1 - 3)`

= `(5/2)^(-2)`

= `(2/5)^2`

= `4/25`

Hence, the first term = t1 = `4/25`

and the common ratio = r = `5/2`.

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अध्याय 2: Sequences and Series - Exercise 2.1 [पृष्ठ २७]

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बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 2 Sequences and Series
Exercise 2.1 | Q 5 | पृष्ठ २७

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