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If for a sequence, tn = nn5n-32n-3, show that the sequence is a G. P. Find its first term and the common ratio. - Mathematics and Statistics

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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.

योग

उत्तर

The sequence (tn) is a G.P. if `("t"_("n" + 1))/"t"_"n"` = constant for all n ∈ N.

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

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

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

= `(5)^(("n" - 2) - ("n" - 3)) xx (2)^(("n" - 3) - ("n" - 2)`

= `(5)^1 xx (2)^(-1)`

= `5/2`

= constant, for all n ∈ N.

∴ The sequence is a G.P. with common ratio (r) = `5/2`

and fiist term = t1 = `(5^(1 - 3))/(2^(1 - 3)`

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

= `2^2/5^2`

= `4/25`

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Sum of the First n Terms of a G.P.
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Sequences and Series - EXERCISE 4.1 [पृष्ठ ५०]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] 11 Standard Maharashtra State Board
अध्याय 4 Sequences and Series
EXERCISE 4.1 | Q 5) | पृष्ठ ५०

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