If in an AP, Pth term is 1/q and qth term is 1/p . Find (pq)th term
Answers
Step-by-step explanation:
As per the question,
pth term of the AP = 1/q
So,
a_{p}=a+(p-1)d.........(1)a
p
=a+(p−1)d.........(1)
Also,
qth term of the AP = 1/p
So,
a_{q}=a+(q-1)d.........(2)a
q
=a+(q−1)d.........(2)
So,
We know that,
(pq)th term of the AP is given by,
a_{pq}=a+(pq-1)d..........(3)a
pq
=a+(pq−1)d..........(3)
Now,
On subtracting eqn. (2) from eqn. (1), we get,
\begin{lgathered}a+(p-1)d=\frac{1}{q}\\and,\\a+(q-1)d=\frac{1}{p}\\(pd-d)-(qd-d)=\frac{1}{q}-\frac{1}{p}\\(p-q)d=\frac{p-q}{pq}\\d=\frac{1}{pq}\end{lgathered}
a+(p−1)d=
q
1
and,
a+(q−1)d=
p
1
(pd−d)−(qd−d)=
q
1
−
p
1
(p−q)d=
pq
p−q
d=
pq
1
Therefore, on putting the value of 'd' in eqn. (1), we get,
\begin{lgathered}a+(p-1)\frac{1}{pq}=\frac{1}{q}\\a+\frac{1}{q}-\frac{1}{pq}=\frac{1}{q}\\a=\frac{1}{pq}\end{lgathered}
a+(p−1)
pq
1
=
q
1
a+
q
1
−
pq
1
=
q
1
a=
pq
1
Therefore, on putting the values of 'a' and 'd' in the eqn.(3), we get,
\begin{lgathered}a_{pq}=a+(pq-1)d\\a_{pq}=\frac{1}{pq}+(pq-1)\frac{1}{pq}\\a_{pq}=\frac{1}{pq}+1-\frac{1}{pq}\\a_{pq}=1\end{lgathered}
a
pq
=a+(pq−1)d
a
pq
=
pq
1
+(pq−1)
pq
1
a
pq
=
pq
1
+1−
pq
1
a
pq
=1
Therefore, we can see that the final evaluated value of the term to be find out is = 1.
Therefore,
a_{pq}=1a
pq
=1
Hence, Proved.
plz refer to this attachment
