Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If a, b, c, d, e are positive real numbers such that a+b+c+d+e=15 and a b2 c3 d4 e5= (120)3 ⋅ 50, then find the value of a2+b2+c2+d2-e2.
Q. If
a
,
b
,
c
,
d
,
e
are positive real numbers such that
a
+
b
+
c
+
d
+
e
=
15
and
a
b
2
c
3
d
4
e
5
=
(
120
)
3
⋅
50
, then find the value of
a
2
+
b
2
+
c
2
+
d
2
−
e
2
.
269
140
Sequences and Series
Report Error
Answer:
5
Solution:
Θ
15
a
+
2
b
+
2
b
+
3
c
+
3
c
+
3
c
+
4
d
+
4
d
+
4
d
+
4
d
+
5
e
+
5
e
+
5
e
+
5
e
+
5
e
=
1
∴
A.M.
=
1
and
G.M.
=
(
a
⋅
2
b
⋅
2
b
⋅
3
c
⋅
3
c
⋅
3
c
⋅
4
d
⋅
4
d
⋅
4
d
⋅
4
d
⋅
5
e
⋅
5
e
⋅
5
e
⋅
5
e
⋅
5
e
)
15
1
=
1
Θ
A.M.
=
G.M.
∴
a
=
2
b
=
3
c
=
4
d
=
5
e
⇒
a
=
1
,
b
=
2
,
c
=
3
,
d
=
4
,
e
=
5
∴
a
2
+
b
2
+
c
2
+
d
2
−
e
2
=
5