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Sum-of-three-numbers-in-GP-be-14-If-one-is-added-to-first-and-second-and-1-is-subtracted-from-the-third-the-new-numbers-are-in-AP-The-smallest-of-them-is-

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Question Number 12827 by 786786AM last updated on 03/May/17
Sum of three numbers in GP be 14. If one is added to first and second and 1 is subtracted from the third, the new numbers are in AP. The smallest of them is
SumofthreenumbersinGPbe14.Ifoneisaddedtofirstandsecondand1issubtractedfromthethird,thenewnumbersareinAP.Thesmallestofthemis
Answered by mrW1 last updated on 04/May/17
a_1 =a a_2 =aq a_3 =aq^2 a_1 +a_2 +a_3 =14 a(1+q+q^2 )=14 (a_2 +1)−(a_1 +1)=(a_3 −1)−(a_2 +1) 2(a_2 +1)=(a_1 +1)+(a_3 −1) 2(aq+1)=a+1+aq^2 −1 aq^2 −2aq+a−2=0 aq^2 +aq+a−3aq−2=0 14−3aq−2=0 aq=4 a+aq+aq^2 =a+4+((16)/a)=14 a+((16)/a)=10 a^2 −10a+16=0 a=((10±(√(100−4×16)))/2)=((10±6)/2)=8 or 2 q=(1/2) or 2 the numbers are 2, 4, 8 or 8, 4, 2 ⇒the smallest of them is 2.
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
the other way, we can start from AP AP : (a−b), a , (a+b) GP : (a−b−1) , (a−1) , (a+b+1) (a−b−1)+(a−1)+(a+b+1)=14 3a−1=14 3a=15 ⇒ a=5 GP : (4−b) , 4 , (6+b) (4−b)(6+b)=4^2 24−2b−b^2 =16 b^2 +2b−8=0 (b+4)(b−2)=0 b=−4 ⇒ 8,4,2 b=2 ⇒2,4,8
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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