find five number in G.P, such that there product is 1024 and 5^th term is square of third term given that term are positive .
Answers
Answer:
4 8 12...............512.
Answer:
The required terms are
1, 2, 4, 8, 16
1,-2, 4, -8, 16
Step-by-step explanation:
Concept used:
The n th term of G.P is t_n=ar^{n-1}tn=arn−1
Let the G.P be \bold{a,ar,ar^2,ar^3,ar^4.........}a,ar,ar2,ar3,ar4.........
Given:
Product of first five terms =1024
\implies\:a.ar.ar^2.ar^3.ar^4=1024⟹a.ar.ar2.ar3.ar4=1024
\implies\:a^5r^{10}=1024⟹a5r10=1024
\implies\:(ar^2)^5=2^{10}⟹(ar2)5=210
\implies\:(ar^2)^5=(2^2)^5⟹(ar2)5=(22)5
\implies\:ar^2=2^2⟹ar2=22
\implies\:ar^2=4⟹ar2=4 ........(1)
Also
t_5=t_3^2t5=t32
\implies\:ar^4=(ar^2)^2⟹ar4=(ar2)2
\implies\:ar^4=a^2r^4⟹ar4=a2r4
\implies\:a=1⟹a=1
Put a=1 in (1) we get
\implies\:r^2=4⟹r2=4
tex]\implies\:r=2,-2[/tex]
When r=2, The 5 terms are
[tex]a=1[\tex]
[tex]ar=1(2)=2[\tex]
[tex]ar^2=1(4)=4[\tex]
[tex]ar^3=1(8)=8[\tex]
[tex]ar^4=1(16)=16[\tex]
When r= - 2, The 5 terms are
[tex]a=1[\tex]
[tex]ar=1(-2)=-2[\tex]
[tex]ar^2=1(4)=4[\tex]
[tex]ar^3=1(-8)=-8[\tex]
[tex]ar^4=1(16)=16[\tex]