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KEAM 2016 Math Paper
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© examsnet.com
Question : 20
Total: 120
If
a
1
,
a
2
,
a
3
,
a
4
are in A.P., then
1
√
a
1
√
a
2
+
1
√
a
2
√
a
3
+
1
√
a
3
√
a
4
=
√
a
1
−
√
a
2
a
3
−
a
2
a
4
−
a
1
a
3
−
a
2
a
3
−
a
2
√
a
4
−
√
a
1
a
1
−
a
4
a
3
−
a
1
a
5
−
a
o
a
1
−
a
4
Validate
Solution:
1
√
a
1
+
√
a
2
+
1
√
a
2
+
√
a
3
+
1
√
a
3
+
√
a
4
=
√
a
1
−
√
a
2
(
√
a
1
+
√
a
2
)
(
√
a
1
−
√
a
2
)
+
√
a
2
−
√
a
3
(
√
a
2
+
√
a
3
)
(
√
a
2
−
√
a
3
)
+
√
a
3
−
√
a
4
(
√
a
3
+
√
a
4
)
(
√
a
3
−
√
a
4
)
[by rationalization]
=
√
a
1
−
√
a
2
a
1
−
a
2
+
√
a
2
−
√
a
3
a
2
−
a
3
+
√
a
3
−
√
a
4
a
3
−
a
4
∵
a
1
,
a
2
,
a
3
and
a
4
are in
A
P
.
∴
a
2
−
a
1
=
a
3
−
a
2
=
a
4
−
a
3
or
a
1
−
a
2
=
a
2
−
a
3
=
a
3
−
a
4
Thus,
1
√
a
1
+
√
a
2
+
1
√
a
2
+
√
a
3
+
1
√
a
3
+
√
a
4
=
√
a
1
−
√
a
2
a
1
−
a
2
+
√
a
2
−
√
a
3
a
1
−
a
2
+
√
a
3
−
√
a
4
a
1
−
a
2
=
(
√
a
1
−
√
a
2
)
+
(
√
a
2
−
√
a
3
)
+
(
√
a
3
−
√
a
4
)
(
a
1
−
a
2
)
=
−
√
a
4
+
√
a
1
a
1
−
a
2
=
√
a
4
−
√
a
1
a
2
−
a
1
=
√
a
4
−
√
a
1
a
3
−
a
2
© examsnet.com
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