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KEAM 2011 Math Paper
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© examsiri.com
Question :
92 of 120
Marks:
+1
,
-0
If
a
1
,
a
2
,
a
3
,
.
.
.
,
a
n
are in A.P. and
a
1
=
0
,
then the value of
(
a
3
a
2
+
a
4
a
3
+
.
.
.
+
a
n
a
n
−
1
)
−
a
2
(
1
a
2
+
1
a
3
+
.
.
.
+
1
a
n
−
2
)
is equal to
(
n
−
2
)
+
1
n
−
2
1
n
−
2
(
n
−
2
)
n
−
1
n
+
2
Validate
Solution:
Given;
a
1
,
a
2
,
a
3
,
ldots
,
a
n
are in
mathrm
‌
A
P
and
a
1
=
0
Then,
a
2
=
a
1
+
d
=
0
+
d
=
d
a
3
=
a
1
+
2
d
=
0
+
2
d
=
2
d
...................................................
....................................................
a
n
=
a
1
+
(
n
−
1
)
d
=
(
n
−
1
)
d
Now,
(
‌
a
3
a
2
+
‌
a
4
a
3
+
.
.
.
+
‌
a
n
a
n
−
1
)
−
a
2
(
‌
1
a
2
+
‌
1
a
3
+
.
.
.
+
‌
1
a
n
−
2
)
=
(
‌
2
d
d
+
‌
3
d
2
d
+
.
.
.
+
‌
(
n
−
1
)
d
(
n
−
2
)
d
)
−
d
(
‌
1
d
+
‌
1
2
d
+
.
.
.
+
‌
1
(
n
−
3
)
d
)
=
(
‌
2
1
+
‌
3
2
+
.
.
.
+
‌
(
n
−
1
)
(
n
−
2
)
)
−
(
1
+
‌
1
2
+
.
.
.
+
‌
1
n
−
3
)
=
{
(
1
+
1
)
+
(
1
+
‌
1
2
)
+
.
.
.
+
(
1
+
‌
1
n
−
2
)
}
−
{
1
+
‌
1
2
+
.
.
.
+
‌
1
n
−
3
}
=
{
(
1
+
1
+
1
+
.
.
.
+
(
n
−
2
)
‌
terms
)
+
(
1
+
‌
1
2
+
‌
1
3
+
.
.
.
+
‌
1
n
−
3
)
+
‌
1
n
−
2
}
−
{
1
+
‌
1
2
+
.
.
.
+
‌
1
n
−
3
}
=
{
(
1
+
1
+
.
.
.
+
(
n
−
2
)
terms
)
+
‌
1
n
−
2
}
=
(
n
−
2
)
+
‌
1
(
n
−
2
)
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