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Byju's Answer
Standard XII
Mathematics
Definition of Relations
100.Given tha...
Question
100.Given that x, a1, a2, y are in A.P. and x, b1, b2, b3, y are also in A.P., then the value of (a2-a1) /(b2-b3) if (x≠ y) (A) 1/2 (B) 3/4 (C) 4/3 (D) 1
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Q.
Assertion :Suppose four distinct positive numbers
a
1
,
a
2
,
a
3
,
a
4
are in GP. Let
b
1
=
a
1
,
b
2
=
b
1
+
a
2
,
b
3
=
b
2
+
a
3
and
b
4
=
b
3
+
a
4
.
STATEMENT- 1: The numbers
b
1
,
b
2
,
b
3
,
b
4
are neither in AP nor in GP, and Reason: STATEMENT- 1: The numbers
b
1
,
b
2
,
b
3
,
b
4
are in HP.
Q.
If
2
y
−
x
=
8
;
5
y
−
x
=
14
,
y
−
2
x
=
1
intersect at
(
a
1
,
b
1
)
,
(
a
2
,
b
2
)
,
(
a
3
,
b
3
)
Find
(
a
1
+
a
2
+
a
3
+
b
1
+
b
2
+
b
3
)
Q.
Suppose four distinct positive numbers
a
1
,
a
2
,
a
3
,
a
4
are in
G
.
P
. Let
b
1
=
a
1
,
b
2
=
b
1
+
a
2
,
b
3
=
b
2
+
a
3
and
b
4
=
b
3
+
a
4
.
STATEMENT-1 : The numbers
b
1
,
b
2
,
b
3
,
b
4
are neither in
A
.
P
. nor in
G
.
P
.
STATEMENT-2 : The numbers
b
1
,
b
2
,
b
3
,
b
4
are in
H
.
P
.
Q.
Suppose four distinct positive number
a
1
,
a
2
,
a
3
,
a
4
are in G.P. Let
b
1
=
a
1
,
b
2
=
b
1
+
a
2
,
a
3
=
b
2
+
a
3
and
b
4
=
b
3
+
a
4
Statement 1 - The numbers
b
1
,
b
2
,
b
3
,
b
4
are neither in A.P. nor in G.P.
Statement 2 - The numaber
b
1
,
b
2
,
b
3
,
b
4
are in H.P.
Q.
If
a
1
,
a
2
,
a
3
,
b
1
,
b
2
,
b
3
∈
R
and are such that
a
i
b
j
≠
1
for
1
≤
i
,
j
≤
3
, then
∣
∣ ∣ ∣ ∣ ∣ ∣
∣
1
−
a
3
1
b
3
1
1
−
a
1
b
1
1
−
a
3
1
b
3
2
1
−
a
1
b
2
1
−
a
3
1
b
3
3
1
−
a
1
b
3
1
−
a
3
2
b
3
1
1
−
a
2
b
1
1
−
a
3
2
b
3
2
1
−
a
2
b
2
1
−
a
3
2
b
3
3
1
−
a
2
b
3
1
−
a
3
3
b
3
1
1
−
a
3
b
1
1
−
a
3
3
b
3
2
1
−
a
3
b
2
1
−
a
3
3
b
3
3
1
−
a
3
b
3
∣
∣ ∣ ∣ ∣ ∣ ∣
∣
> 0 Provided either
a
1
<
a
2
<
a
3
and
b
1
<
b
2
<
b
3
, or
a
1
>
a
2
a
3
and
b
1
>
b
2
>
b
3
then show
(
a
1
−
a
2
)
(
a
2
−
a
3
)
(
a
3
−
a
1
)
(
b
1
−
b
2
)
(
b
2
−
b
3
)
(
b
3
−
b
1
)
<
0
,
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