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Question 3 (viii)
In an AP:
(viii) Given an=4, d = 2, Sn=−14, find n and a.
In an AP:
(viii) Given an=4, d = 2, Sn=−14, find n and a.
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Solution
(viii)
Given that, an=4,d=2,Sn=−14an=a+(n−1)d4=a+(n−1)24=a+2n−2a+2n=6a=6−2n……(i)Sn=n2(a+an)−14=n2(a+4)−28=n(a+4)−28=n(6−2n+4) From equation (i)−28=n(−2n+10)−28=−2n2+10n2n2−10n−28=0n2−5n−14=0n(n−7)+2(n−7)=0(n−7)(n+2)=0Either n−7=0 or n+2=0n=7 or n=−2However, n can neither be negative nor fractional.Therefore, n=7From equation (i), we geta=6−2na=6−2(7)=6−14=−8
Given that, an=4,d=2,Sn=−14an=a+(n−1)d4=a+(n−1)24=a+2n−2a+2n=6a=6−2n……(i)Sn=n2(a+an)−14=n2(a+4)−28=n(a+4)−28=n(6−2n+4) From equation (i)−28=n(−2n+10)−28=−2n2+10n2n2−10n−28=0n2−5n−14=0n(n−7)+2(n−7)=0(n−7)(n+2)=0Either n−7=0 or n+2=0n=7 or n=−2However, n can neither be negative nor fractional.Therefore, n=7From equation (i), we geta=6−2na=6−2(7)=6−14=−8
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