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nscuivebddIhlegers,sumofwhosesquaresis290.18. How man419. If tangents PA and PR from o

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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