how many terms of the ap 9 17 25 must be taken to give a sum of 636
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Answered by
14
sn=n/2 (2a+(n-1)d)
636=n/2 (2 (9)+(n-1)8)
636×2=n (18+8n-8)
1272=10n+8n^2
8n^2+10n-1272=0÷2
4n^2+5n-636=0
4n^2+53n-48n-636=0
n (4n+53)-12 (4n-53)=0
(4n+53)(n-12)=0
n-12=0 4n+53=0
n=12 n=-53/4
n=-53/4 is discarded
12 terms of the ap 9 17 25 must be taken to give a sum of 636
sofiyasofii244:
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Answered by
16
Suppose first term be a
Common Difference = 17 - 9 = 8
Sum is 636
Hence
⇒ n(4n + 5) = 636
⇒ 4n^2 + 5n + 636 = 0
In this situation “n” cannot be negative
n = 12
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