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KEAM 2017 Math Paper
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© examsnet.com
Question : 32
Total: 120
Let a, a + r and a + 2r be positive real numbers such that their product is 64. Then the minimum value of a + 2r is equal to
4
3
2
1/2
1
Validate
Solution:
We know AM
≥
G
M
a
+
(
a
+
r
)
+
(
a
+
2
r
)
3
≥
(
a
(
a
+
r
)
(
a
+
2
r
)
)
1
3
⇒
3
(
a
+
r
)
3
≥
(
64
)
1
3
⇒
(
a
+
r
)
≥
4
Also,
64
=
a
(
a
+
r
)
(
a
+
2
r
)
⇒
64
≥
(
4
−
r
)
×
4
(
r
+
4
)
⇒
16
≥
16
−
r
2
⇒
r
2
≤
0
∴
r
=
0
Now,
a
+
2
r
=
4
+
0
=
4
© examsnet.com
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