मराठी

Find the value of n so that an+1+bn+1an+bn may be the geometric mean between a and b. - Mathematics

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प्रश्न

Find the value of n so that  `(a^(n+1) + b^(n+1))/(a^n + b^n)` may be the geometric mean between a and b.

बेरीज

उत्तर

The geometric mean between a and b = `sqrt"ab"`

⇒ `("a"^("n"+ 1) + "b"^("n" + 1))/("a"^"n" + "b"^"n") = sqrt"ab"`

∴ `"a"^("n"+ 1) + "b"^("n" + 1) = sqrt"ab" ("a"^"n" + "b"^"n")` 

= `"a"^("n"+ 1/2) "b"^(1/2) + "a"^(1/2) "b"^("n" + 1/2)`

or `("a"^("n" + 1) - "a"^("n" + 1/2) "b"^(1/2)) - ("a"^(1/2) "b"^("n" + 1/2) - "b"^("n" + 1)) = 0`

or `"a"^("n" + 1/2) ("a"^(1/2) - "b"^(1/2)) - "b"^ ("n" + 1/2)("a" ^(1/2) - "b"^(1/2)) = 0`

or `("a"^(1/2) - "b"^(1/2)) ("a"^("n" + 1/2) - "b"^ ("n" + 1/2)) = 0`

`"a" ^(1/2) - "b"^(1/2) ≠ 0`

∴ `"a"^("n" + 1/2) - "b"^ ("n" + 1/2) = 0`

or `"a"^("n" + 1/2) = "b"^ ("n" + 1/2)`

or `("a"/"b")^("n"+1/2) = 1 = ("a"/"b")^0`

⇒ `"n"+ 1/2 = 0` 

n = `(-1)/2`

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पाठ 9: Sequences and Series - Exercise 9.3 [पृष्ठ १९३]

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एनसीईआरटी Mathematics [English] Class 11
पाठ 9 Sequences and Series
Exercise 9.3 | Q 27 | पृष्ठ १९३

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