Question

if pth term of an ap is 1/q and qth term is 1/p then prove that (pq)th term is 1

Answer 1

hav u got D answer....here I hav put eqn. 1 and 2 in a+(pq-1)d

Answer 2

Answer:

As per the question,

pth term of the AP = 1/q

So,

$a_{p}=a+(p-1)d.........(1)$

Also,

qth term of the AP = 1/p

So,

$a_{q}=a+(q-1)d.........(2)$

So,

We know that,

(pq)th term of the AP is given by,

$a_{pq}=a+(pq-1)d..........(3)$

Now,

On subtracting eqn. (2) from eqn. (1), we get,

$a+(p-1)d=\frac{1}{q}\\and,\\a+(q-1)d=\frac{1}{p}\\(pd-d)-(qd-d)=\frac{1}{q}-\frac{1}{p}\\(p-q)d=\frac{p-q}{pq}\\d=\frac{1}{pq}$

Therefore, on putting the value of 'd' in eqn. (1), we get,

$a+(p-1)\frac{1}{pq}=\frac{1}{q}\\a+\frac{1}{q}-\frac{1}{pq}=\frac{1}{q}\\a=\frac{1}{pq}$

Therefore, on putting the values of 'a' and 'd' in the eqn.(3), we get,

$a_{pq}=a+(pq-1)d\\a_{pq}=\frac{1}{pq}+(pq-1)\frac{1}{pq}\\a_{pq}=\frac{1}{pq}+1-\frac{1}{pq}\\a_{pq}=1$

Therefore, we can see that the final evaluated value of the term to be find out is = 1.

Therefore,

$a_{pq}=1$

Hence, Proved.