if (p+q)th term of an A.P is m and (p-q)th term is n, then pth term is what
Answers
Answer:
Step-by-step explanation:
umbers.
+=,−=
t
p
+
q
=
m
,
t
p
−
q
=
m
(,∈ℕ
p
,
q
∈
N
)
The most likely answer is
m
, and I will prove this for both GPs and APs. Other sequences do exist, or we could take the sequence as a function ()
f
(
x
)
, where there would be an infinite number of solutions for the
p
th term.
First, GP:
=−1
t
n
=
a
r
n
−
1
Let =+
Let
n
=
p
+
q
+==+−1
t
p
+
q
=
m
=
a
r
p
+
q
−
1
Similarly,
−==−−1
t
p
−
q
=
m
=
a
r
p
−
q
−
1
=−1
t
p
=
a
r
p
−
1
We can multiply +
t
p
+
q
and −
t
p
−
q
together:
+−=+−
t
p
+
q
t
p
−
q
=
t
p
+
q
t
p
−
q
On the RHS, let +,−=
t
p
+
q
,
t
p
−
q
=
m
and on the LHS, let +=+−1,−=−−1
t
p
+
q
=
a
r
p
+
q
−
1
,
t
p
−
q
=
a
r
p
−
q
−
1
+−1−−1=2
a
r
p
+
q
−
1
a
r
p
−
q
−
1
=
m
2
2+−1+−−1=2
a
2
r
p
+
q
−
1
+
p
−
q
−
1
=
m
2
22−2=2
a
2
r
2
p
−
2
=
m
2
22(−1)=2
a
2
r
2
(
p
−
1
)
=
m
2
(−1)2=2
(
a
r
p
−
1
)
2
=
m
2
−1=±
a
r
p
−
1
=
±
m
As −1=
a
r
p
−
1
=
t
p
=±
t
p
=
±
m
Now for an AP (which will have less explaining, as most of the explanations above apply here):
=+(−1)
t
n
=
a
+
d
(
n
−
1
)
+==+(+−1)
t
p
+
q
=
m
=
a
+
d
(
p
+
q
−
1
)
−==+(−−1)
t
p
−
q
=
m
=
a
+
d
(
p
−
q
−
1
)
=+(−1)
t
p
=
a
+
d
(
p
−
1
)
We can add +
t
p
+
q
and −
t
p
−
q
together and substitute their respective definitions
+(+−1)++(−−1)=2
a
+
d
(
p
+
q
−
1
)
+
a
+
d
(
p
−
q
−
1
)
=
2
m
++−++−−=2
a
+
d
p
+
d
q
−
d
+
a
+
d
p
−
d
q
−
d
=
2
m
2+2−2=2
2
a
+
2
d
p
−
2
d
=
2
m
2(+−)=2
2
(
a
+
d
p
−
d
)
=
2
m
+−=
a
+
d
p
−
d
=
m
+(−1)=
a
+
d
(
p
−
1
)
=
m
=