the sum of 3 numbers in ap is 18 and their product is 192 .find the numbers
Answers
Answer:
The numbers are 4,6,8
Step-by-step explanation:
Let the numbers be a-d,a,a+d
Sum of numbers=3a=18
a=6
Product of numbers=a(a+d)(a-d)
=a(a^2-d^2)=192
6(36-d^2)=192
36-d^2=32
d^2=4
d=2
The numbers are a-d,a a+d
The numbers are 4,6,8
Answer:
\begin{lgathered}Required\:three \: number\\are \:(4,6,8)\:Or\:(8,6,4)\end{lgathered}
Requiredthreenumber
are(4,6,8)Or(8,6,4)
Step-by-step explanation:
\begin{lgathered}Let \: (a-d),a,(a+d)\:are \\three \: consecutive\: terms\\of \:A.P\end{lgathered}
Let(a−d),a,(a+d)are
threeconsecutiveterms
ofA.P
According to the problem given,
sum \:of \: numbers=18sumofnumbers=18
\implies a-d+a+a+d=18⟹a−d+a+a+d=18
\implies 3a=18⟹3a=18
\implies a = \frac{18}{3}=6⟹a=
3
18
=6
Product\:of \: numbers=192Productofnumbers=192
\implies (a-d)a(a+d)=192⟹(a−d)a(a+d)=192
\implies (a^{2}-d^{2}a=192⟹(a
2
−d
2
a=192
\implies (6^{2}-d^{2})\times 6=192⟹(6
2
−d
2
)×6=192
\implies 36-d^{2}=\frac{192}{6}⟹36−d
2
=
6
192
\implies 36-d^{2}=32⟹36−d
2
=32
\implies -d^{2}=32-36⟹−d
2
=32−36
\implies -d^{2}=-4⟹−d
2
=−4
\implies d^{2}=4⟹d
2
=4
\implies d = ±\sqrt{2^{2}}⟹d=±
2
2
\implies d = ± 2⟹d=±2
\begin{lgathered}Case\: 1 \\If \:a = 6,\:d=2\end{lgathered}
Case1
Ifa=6,d=2
\begin{lgathered}The\: numbers\:are ,\\(a-d),a,(a+d)\\(6-2),6,(6+2)\\4,6,8\end{lgathered}
Thenumbersare,
(a−d),a,(a+d)
(6−2),6,(6+2)
4,6,8
\begin{lgathered}Case \:2 \\If \:a=6,\:d=-2\\Three