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JEE Advanced 2016 Paper 2
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© examsnet.com
Question : 40
Total: 54
Let
b
i
>
1
for
i
=
1
,
2
,
.
.
.
.
.
,
101
. Suppose
l
o
g
e
b
1
,
l
o
g
e
b
2
,
.
.
.
.
.
,
l
o
g
e
b
101
are in Arithmetic Progression (A.P.) with the common difference
l
o
g
e
2
. Suppose
a
1
,
a
2
,
.
.
.
.
.
.
,
a
101
are in A.P. such that
a
1
=
b
1
and
a
51
=
b
51
. If
t
=
b
1
+
b
2
+
.
.
.
.
.
+
b
51
and
s
=
a
1
+
a
2
+
.
.
.
.
+
a
51
then
s
>
t
and
a
101
>
b
101
s
>
t
and
a
101
<
b
101
s
<
t
and
a
101
>
b
101
s
<
t
and
a
101
<
b
101
Validate
Solution:
If
log
e
b
1
,
log
e
b
2.
.
.
.
log
e
b
101
→
A
P
;
D
=
log
e
2
⇒
b
1
b
2
b
3.
.
.
.
b
101
→
G
P
;
r
=
2
∴
b
1
,
2
b
1
,
2
2
b
1
,
.
.
.
.
.
,
2
100
b
1
,
.
.
.
.
.
.
G
P
a
1
a
2
a
3.
.
.
.
.
.
.
a
101.
.
.
.
.
.
.
A
P
Given,
a
1
=
b
1
&
a
51
=
b
51
⇒
a
1
+
50
D
=
2
50
b
1
∴
a
1
=
50
D
=
2
50
a
1
(
A
s
b
1
=
a
1
)
Now,
t
=
b
1
(
2
51
−
1
)
;
s
=
51
2
(
2
a
1
+
50
D
)
t
=
a
1
.2
51
−
a
1
⇒
t
<
a
1
.2
51.
.
.
.
.
(
i
)
;
s
=
51
2
(
a
1
+
a
1
+
50
D
)
s
=
51
2
(
a
1
+
2
50
a
1
)
s
=
51
a
1
2
+
51
2
.2
50
a
1
s
>
a
1
.2
51
.......(ii)
clearly s > t (from equation (i) and (ii))
Also
a
101
=
a
1
+
100
D
;
b
101
=
b
1
.2
100
∴
a
101
=
a
1
+
100
(
2
50
a
1
−
a
1
50
)
;
b
101
=
2
100
a
1
.....(iii)
a
101
=
a
1
+
2
51
a
1
−
2
a
1
⇒
a
101
=
2
51
a
1
−
a
1
⇒
a
101
<
2
51
a
1
......(iv)
clearly
b
101
>
a
101
(from equation (iii) and (iv))
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