Question

if pth term of an AP is 1/q and qth term is 1/p then pqth term is detail explanation required and i wil mark as brainliest

Answer 1

Answer:

Let the first term be a and common difference be d.

According to question:

= > pth term = 1/q

= > a + (p-1)d = 1/q

= > a + pd - d = 1/q ...(1)

= > qth term = 1/p

= > a + qd - d = 1/p ...(2)

(2) - (1):

= > qd - pd = 1/p - 1/q

= > d(q-p) = (q-p)/pq

= > d = 1/pq

Hence,

= > a + pd - d = 1/q

= > a + p(1/pq) - (1/pq ) = 1/q

= > a = 1/pq = d

Therefore,

= > pdth term

= > a + ( pq - 1 )d

= > d + ( pq -1 )d

= > d + pqd - d

= > pqd

= > pq(1/pq)

= > 1

Hence the pqth term is 1.

Answer 2

$\huge\star \: {\sf{\underline{\red{Given}}}}$

Let the first term be a and the difference be d.

$\huge{\bf{\underline{\red{A. T. Q, }}}}$

$\implies{\tt{\pink{pth term = \frac{1}{q}}}}$

$\implies{\tt{\pink{pth term = \frac{1}{q}}}}$

$\implies{\tt{\pink{a+(p-1)d = \frac{1}{q}}}}$

$\implies{\tt{\pink{ a+pd - d = \frac{1}{q}------(1)}}}$

$\implies{\tt{\pink{qth term = \frac{1}{p}}}}$

$\implies{\tt{\pink{a + qd - d = \frac{1}{p}-----(2)}}}$

Here we will subtract the eq--(2) - eq(1),

$\implies{\tt{\green{qd - pd = \frac{1}{p} - \frac{1}{q}}}}$

$\implies{\tt{\green{d(q - p) = \frac{q - p}{pq} }}}$

$\implies{\tt{\green{d = \frac{1}{pq}}}}$

HenCe,

$\implies{\tt{\green{a + pd - d = \frac{1}{q}}}}$

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

$\implies{\tt{\green{\frac{ a = 1}{pq = d}}}}$

Therefore,

$\implies{\tt{\green{pdth \: term}}}$

$\implies{\tt{\green{a + (pq-1)d}}}$

$\implies{\tt{\green{d +(pq - 1)d}}}$

$\implies{\tt{\green{d +pqd -d}}}$

$\implies{\tt{\green{pq \frac{1}{pq}}}}$

$\implies{\tt{\green{1}}}$

${\tt{\pink{\therefore{The \: pqth \: term = 1}}}}$