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inu1weconsecutiveoddintegers,290sHow many terms to the A.P. : 9, 17,25,.... . must be taken to give a sumsum of whose squares isof 636? 419. If tangents PA and PB from a noint Pto cirgh centreare inclined to each other at |
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Answer» A.p= 9,17,25a=9d=17-9=8Sum of n terms=n÷2[2a+(n-1)d 636=n÷2[2(9)+(n-1)8]636=n÷2[18+8n-8]636=n÷2[10+8n]1272=10n+8n^28n^2+10n-1272=02[4n^2+5n-636]=04n^2+5n-636=04n^2+53n-48n-636=0n(4n+53)-12(4n+53)=0(n-12)(4n+53)=0n-12=0n=12 12 terms must be given to sum of 636 |
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