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11TH · MATH

Question 2

menu_bookSolutionmenu_bookTheorymenu_book•(i)an=3n+5a_n = 3n + 5an​=3n+5menu_book•(ii)an+1=4an−7a_{n+1} = 4a_n - 7an+1​=4an​−7and $a_1 = 3$menu_book•(iii)an=(n−3)(n+1)a_n = (n-3)(n+1)an​=(n−3)(n+1)menu_book•(iv)a1=−1a_1 = -1a1​=−1, $a_{n+1} = \dfrac{3}{a_n + 2}$menu_book•(v)an=8−203+na_n = 8 - \dfrac{20}{3+n}an​=8−3+n20​menu_book•(vi)a1=1a_1 = 1a1​=1, $a_{n+1} = (3a_n + 2)^2$menu_book•(vii)an=(−2n)2a_n = (-2n)^2an​=(−2n)2menu_book•(viii)an=(−1)n7n2a_n = (-1)^n 7n^2an​=(−1)n7n2
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11TH · MATH

Question 2

menu_bookSolutionmenu_bookTheorymenu_book•(i)an=3n+5a_n = 3n + 5an​=3n+5menu_book•(ii)an+1=4an−7a_{n+1} = 4a_n - 7an+1​=4an​−7and $a_1 = 3$menu_book•(iii)an=(n−3)(n+1)a_n = (n-3)(n+1)an​=(n−3)(n+1)menu_book•(iv)a1=−1a_1 = -1a1​=−1, $a_{n+1} = \dfrac{3}{a_n + 2}$menu_book•(v)an=8−203+na_n = 8 - \dfrac{20}{3+n}an​=8−3+n20​menu_book•(vi)a1=1a_1 = 1a1​=1, $a_{n+1} = (3a_n + 2)^2$menu_book•(vii)an=(−2n)2a_n = (-2n)^2an​=(−2n)2menu_book•(viii)an=(−1)n7n2a_n = (-1)^n 7n^2an​=(−1)n7n2
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Accedmychevron_right11thchevron_rightmathchevron_rightSequences And Serieschevron_rightExercise 6.1

See solution.mdx for the answer.

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