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How many terms of the AP: 9, 17, 25, . . . must be taken to give a sum of 636? |
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Answer» d=8 a=9 `S_n=n/2(2a+(n-1)d)` 636=n/2(2*9+(n-1)8) `4n^2+5n-636` n(4n+53)-12(4n+53)=0 (n-12)(4n+53)=0 n=12,-53/4 n can not be negative so, n=12 |
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