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(vi)given a = 4, d=2, S =-14, find n and a.tolo |
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Answer» Sn = n/2(2a+(n+1)d) Now an= a+(n-1)d4= a+(n-1)a+n = 5 An=a+(n-1)d=4a+(n-1)2=4 (substitute d=2)a+2n-2=4a+2n=6a=6-2n Sn=n/2(2a+(n-1)d)-14=n/2(2(6-2n)+(n-1)2) (substitute a=6-2n,d=2)-14=n/2(12-4n+2n-2)-14=n/2(10-2n)-14=n(5-n)5n-n^2=-14n^2-5n-14=0n^2-7n+2n-14=06-2n(n-7)+2(n-7)=0(n-7)(n+2)=0n=7,-2 n can't be negativetherefore n=7 a=6-2n (substitute n=7)a=6-2(7)=6-14=-8 |
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