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Byju's Answer
Standard XII
Mathematics
nth Term of A.P
Let a1,a2,a...
Question
Let
a
1
,
a
2
,
a
3
,
.
.
.
.
be terms of an A.P. If
a
1
+
a
2
+
.
.
.
.
.
+
a
p
a
1
+
a
2
+
.
.
.
.
.
+
a
q
=
p
2
q
2
,
p
≠
q
, then
a
6
a
21
equals
Open in App
Solution
a
1
+
a
2
+
.
.
.
.
+
a
p
a
1
+
a
2
+
.
.
.
.
+
a
q
=
p
2
q
2
⇒
p
2
[
2
a
1
+
(
p
−
1
)
d
]
q
2
[
2
a
1
+
(
q
−
1
)
d
]
=
p
2
q
2
⇒
2
a
1
+
(
p
−
1
)
d
2
a
1
+
(
q
−
1
)
d
=
p
q
by cancelling of the same terms both sides
⇒
2
a
1
q
+
(
p
−
1
)
q
d
=
2
a
1
p
+
(
q
−
1
)
p
d
by cross multiplication
⇒
2
a
1
(
q
−
p
)
=
[
(
q
−
1
)
p
−
(
p
−
1
)
q
]
d
by grouping the common terms
⇒
−
2
a
1
(
p
−
q
)
=
[
p
q
−
p
−
p
q
+
q
]
d
⇒
−
2
a
1
(
p
−
q
)
=
−
(
p
−
q
)
d
on simplifying
⇒
2
a
1
=
d
.....
(
1
)
So
a
6
a
21
=
a
1
+
5
d
a
1
+
20
d
=
a
1
+
5
(
2
a
1
)
a
1
+
20
(
2
a
1
)
From
(
1
)
=
11
a
1
41
a
1
=
11
41
Suggest Corrections
0
Similar questions
Q.
Let
a
1
,
a
2
,
a
3
.
.
.
.
be terms of an A.P. If
a
1
+
a
2
+
.
.
.
.
+
a
p
a
1
+
a
2
+
.
.
.
.
+
a
q
=
p
2
q
2
,
p
≠
q
, then
a
6
a
21
equals
Q.
Let
a
1
,
a
2
,
a
3
,
.
.
.
.
.
be terms of an
A
.
P
.
if
a
1
+
a
2
+
.
.
.
+
a
q
a
1
+
a
2
+
.
.
.
+
a
q
=
p
2
q
2
,
(
p
≠
q
)
then find
a
6
a
21
.
Q.
Let
a
1
,
a
2
,
a
3
,
.
.
.
.
be terms of an A.P.
If
a
1
+
a
2
+
.
.
.
.
+
a
p
a
1
+
a
2
+
.
.
.
.
.
+
a
q
=
p
2
q
2
,
p
≠
q
then
a
6
a
21
must be-
Q.
Let
a
1
,
a
2
,
a
3
,
…
be terms of an A.P. If
a
1
+
a
2
+
⋯
+
a
p
a
1
+
a
2
+
⋯
+
a
q
=
p
2
q
2
(
p
≠
q
)
, then
a
6
a
21
is equal to
Q.
Let
a
1
,
a
2
,
a
3
⋯
be terms of A.P.
I
f
a
1
+
a
2
+
⋯
a
p
a
1
+
a
2
+
⋯
+
a
q
=
p
2
q
2
,
p
≠
q
,
then
a
6
a
21
equal
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