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JEE Main Math Class 11 Sequences and Series Part 3 Questions
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© examsnet.com
Question : 97
Total: 101
If
1
(
20
−
a
)
(
40
−
a
)
+
1
(
40
−
a
)
(
60
−
a
)
+
.
.
.
+
1
(
180
−
a
)
(
200
−
a
)
=
1
256
, then the maximum value of
a
is :
[29-Jul-2022-Shift-1]
198
202
212
218
Validate
Solution:
1
20
(
1
20
−
a
−
1
40
−
a
+
1
40
−
a
−
1
60
−
a
+
.
.
.
+
1
180
−
a
−
1
200
−
a
)
=
1
256
⇒
1
20
(
1
20
−
a
−
1
200
−
a
)
=
1
256
⇒
1
20
(
180
(
20
−
a
)
(
200
−
a
)
)
=
1
256
⇒
(
20
−
a
)
(
200
−
a
)
=
9.256
OR
a
2
−
220
a
+
1696
=
0
⇒
a
=
212
,
8
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