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How many terms of the A.P.: 9,17,25,..... must be taken to give a sum of 636?

Answer» 9, 17, 25, ....
Here, a=9,
`d=17-9=25-17=8`
Let `S_(n)=636`
`rArr (n)/(2)[2(9)+(n-1)8]=636`
`rArr (n)/(2)(18+8n-8)=636`
`rArr n(4n+5)=636`
`rArr 4n^(2)+5n-636=0`
`rArr (4n+53)(n-12)=0`
`rArr n=-(53)/(4) or n=12`
`n=-(53)/(4)` is not possible
`:. n=12`


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