the first term of a GP is 10 then the product of first five terms ?
Answers
Answer:
Let the 5 terms in geometric progression be
r
2
a
,
r
a
,a,ar,ar
2
Now given that
Product of these numbers =1
i.e.,
r
2
a
×
r
a
×a×ar×ar
2
=1
a
5
=1
a
5
=1
5
⇒ a=1
and
sum of first 3 terms =
4
7
i.e.,
r
2
a
+
r
a
+a=
4
7
r
2
a+ar+ar
2
=
4
7
4(a+ar+ar
2
)=7r
2
But a=1
∴ 4(1+1r+1r
2
)=7r
2
Or, 7r
2
−4r
2
−4r−4=0
3r
2
−4r−4=0
Solve by factorisation method
3r
2
−4r−4=0
3r
2
−6r+2r−4=0
3r(r−2)+2(r−2)=0
(r−2)(3r+2)=0
⇒ r−2=0 or 3r+2=0
r=2 or 3r=−2
r=−
3
2
∴ Common ratio of geometric progression be =r=2 or −
3
2
Step-by-step explanation:
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Answer:
Let the 5 terms in geometric progression be
r
2
a
,
r
a
,a,ar,ar
2
Now given that
Product of these numbers =1
i.e.,
r
2
a
×
r
a
×a×ar×ar
2
=1
a
5
=1
a
5
=1
5
⇒ a=1
and
sum of first 3 terms =
4
7
i.e.,
r
2
a
+
r
a
+a=
4
7
r
2
a+ar+ar
2
=
4
7
4(a+ar+ar
2
)=7r
2
But a=1
∴ 4(1+1r+1r
2
)=7r
2
Or, 7r
2
−4r
2
−4r−4=0
3r
2
−4r−4=0
Solve by factorisation method
3r
2
−4r−4=0
3r
2
−6r+2r−4=0
3r(r−2)+2(r−2)=0
(r−2)(3r+2)=0
⇒ r−2=0 or 3r+2=0
r=2 or 3r=−2
r=−
3
2
∴ Common ratio of geometric progression be =r=2 or −
3
2