how many terms of AP 17 15 13 should be taken so that their sum is 81
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heya! here is ur answer
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pritish9:
thanx
wlcm
nice.......
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a= 17
d= - 2
Sn = 81
=> n/2 [2a +(n-1)d] = 81
=> n[2*17 + (n-1)(-2)] = 162
=> n (34 - 2n +2) = 162
=> n(36 - 2n) = 162
=> 36n - 2n^2 - 162 = 0
=> 2n^y -36n +162 = 0
=> n^2 - 18n + 81 = 0
=> n^2 - 9n - 9n + 81 = 0
=> n(n-9) - 9(n-9) = 0
=> (n-9)(n-9) = 0
=> (n-9)^2 = 0
=> n - 9 = 0
=> n = 9
No. of terms = 9
d= - 2
Sn = 81
=> n/2 [2a +(n-1)d] = 81
=> n[2*17 + (n-1)(-2)] = 162
=> n (34 - 2n +2) = 162
=> n(36 - 2n) = 162
=> 36n - 2n^2 - 162 = 0
=> 2n^y -36n +162 = 0
=> n^2 - 18n + 81 = 0
=> n^2 - 9n - 9n + 81 = 0
=> n(n-9) - 9(n-9) = 0
=> (n-9)(n-9) = 0
=> (n-9)^2 = 0
=> n - 9 = 0
=> n = 9
No. of terms = 9
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