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Question Number 12827 by 786786AM last updated on 03/May/17
SumofthreenumbersinGPbe14.Ifoneisaddedtofirstandsecondand1issubtractedfromthethird,thenewnumbersareinAP.Thesmallestofthemis
Answered by mrW1 last updated on 04/May/17
a1=aa2=aqa3=aq2a1+a2+a3=14a(1+q+q2)=14(a2+1)−(a1+1)=(a3−1)−(a2+1)2(a2+1)=(a1+1)+(a3−1)2(aq+1)=a+1+aq2−1aq2−2aq+a−2=0aq2+aq+a−3aq−2=014−3aq−2=0aq=4a+aq+aq2=a+4+16a=14a+16a=10a2−10a+16=0a=10±100−4×162=10±62=8or2q=12or2thenumbersare2,4,8or8,4,2⇒thesmallestofthemis2.
Answered by sandy_suhendra last updated on 04/May/17
theotherway,wecanstartfromAPAP:(a−b),a,(a+b)GP:(a−b−1),(a−1),(a+b+1)(a−b−1)+(a−1)+(a+b+1)=143a−1=143a=15⇒a=5GP:(4−b),4,(6+b)(4−b)(6+b)=4224−2b−b2=16b2+2b−8=0(b+4)(b−2)=0b=−4⇒8,4,2b=2⇒2,4,8
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