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Question Number 17595 by virus last updated on 08/Jul/17

Let a,b,c be the posetive integer such that b/a is an also integer if a,b,c are in GP and AM of a,b,c is(b+2) then find the value of (a^2 +a−14)/(a+1)

$${Let}\:{a},{b},{c}\:{be}\:{the}\:{posetive}\:{integer}\:{such}\:{that}\: \\ $$$${b}/{a}\:{is}\:{an}\:{also}\:{integer}\:{if}\:{a},{b},{c}\:{are}\:{in}\:{GP}\:{and}\: \\ $$$${AM}\:{of}\:{a},{b},{c}\:{is}\left({b}+\mathrm{2}\right)\:{then}\:{find}\:{the}\:{value}\:{of} \\ $$$$\left({a}^{\mathrm{2}} +{a}−\mathrm{14}\right)/\left({a}+\mathrm{1}\right) \\ $$

Commented by virus last updated on 08/Jul/17

fine

$${fine} \\ $$

Answered by mrW1 last updated on 08/Jul/17

b=na (b/a)=(c/b) ⇒c=n^2 a ((a+b+c)/3)=b+2 ⇒((a+na+n^2 a)/3)=na+2 ⇒a+na+n^2 a=3na+6 ⇒n^2 −2n+(1−(6/a))=0 n=((2+(√(4−4(1−(6/a)))))/2)=1+(√(6/a)) ⇒a=6 ⇒n=2⇒b=12⇒c=24 (a^2 +a−14)/(a+1)=(36+6−14)/7=4

$$\mathrm{b}=\mathrm{na} \\ $$$$\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}} \\ $$$$\Rightarrow\mathrm{c}=\mathrm{n}^{\mathrm{2}} \mathrm{a} \\ $$$$\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{3}}=\mathrm{b}+\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{a}+\mathrm{na}+\mathrm{n}^{\mathrm{2}} \mathrm{a}}{\mathrm{3}}=\mathrm{na}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{na}+\mathrm{n}^{\mathrm{2}} \mathrm{a}=\mathrm{3na}+\mathrm{6} \\ $$$$\Rightarrow\mathrm{n}^{\mathrm{2}} −\mathrm{2n}+\left(\mathrm{1}−\frac{\mathrm{6}}{\mathrm{a}}\right)=\mathrm{0} \\ $$$$\mathrm{n}=\frac{\mathrm{2}+\sqrt{\mathrm{4}−\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{6}}{\mathrm{a}}\right)}}{\mathrm{2}}=\mathrm{1}+\sqrt{\frac{\mathrm{6}}{\mathrm{a}}} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{6} \\ $$$$\Rightarrow\mathrm{n}=\mathrm{2}\Rightarrow\mathrm{b}=\mathrm{12}\Rightarrow\mathrm{c}=\mathrm{24} \\ $$$$\left(\mathrm{a}^{\mathrm{2}} +\mathrm{a}−\mathrm{14}\right)/\left(\mathrm{a}+\mathrm{1}\right)=\left(\mathrm{36}+\mathrm{6}−\mathrm{14}\right)/\mathrm{7}=\mathrm{4} \\ $$

Commented by virus last updated on 08/Jul/17

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by chux last updated on 09/Jul/17

mr W1.... please how is n=2

$$\mathrm{mr}\:\mathrm{W1}....\:\mathrm{please}\:\mathrm{how}\:\mathrm{is}\:\mathrm{n}=\mathrm{2} \\ $$

Commented by mrW1 last updated on 09/Jul/17

n=1+(√(6/a)) since n and a are positive integers, the only solution is a=6 and then n=2.

$$\mathrm{n}=\mathrm{1}+\sqrt{\frac{\mathrm{6}}{\mathrm{a}}} \\ $$$$\mathrm{since}\:\mathrm{n}\:\mathrm{and}\:\mathrm{a}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers}, \\ $$$$\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{a}=\mathrm{6}\:\mathrm{and}\:\mathrm{then} \\ $$$$\mathrm{n}=\mathrm{2}. \\ $$

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