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CAT Exam Model Paper 4 with solutions for free online practice
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© examsnet.com
Question : 9
Total: 60
If S
=
1
100
+
1
101
+
1
102
+
.
.
.
+
1
1000
then
S
≤
1
3
1
3
,
S
≤
2
3
2
3
,
S
≤
1
S > 1
Validate
Solution:
S
=
1
100
+
1
101
+
1
102
+
.
.
.
+
1
998
+
1
999
+
1
1000
=
(
1
100
+
1
1000
)
+
(
1
101
+
1
999
)
+
.
.
.
(
450
terms
)
+
1
550
=
1000
+
100
100
×
1000
+
999
+
101
101
×
999
+
.
.
.
.
+
1
550
=
1100
100
×
1000
+
1100
101
×
999
+
.
.
.
+
1
550
In the first 450 terms, the sum of the two numbers in the denominator is constant. Thus, their product will be maximum when they are equal
∴
S
>
1100
550
×
550
+
1100
550
×
550
+
.
.
.
(
450
terms
)
+
1
550
or
S
>
450
(
1100
550
×
550
)
+
1
550
>
1
© examsnet.com
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