Question

If pth and qth termof an AP are 1/qr and 1/pr respectively, find the rth term

Answer 1

Step-by-step explanation:

$let \: first \: term \: of \: a.p. = a \\ common \: difference = d \\ pth \: term \: = a + (p - 1) \times d = \frac{1}{qr} .......(1) \\ \\ qth \: term \: = a + (q - 1) \times d = \frac{1}{pr} ......(2) \\ \\ substracting \: (2) \: from \: (1) \\ (p - 1 - q + 1) \times d = \frac{(p - q)r}{pq {r}^{2} } \\ d = \frac{1}{pqr} \\ \\ substuiting \: the \: value \: of \: d \: in \: (1) \\ we \: get \: \: \: a = \frac{1}{pqr} \\ \\ rth \: term \: = a + (r - 1) \times d \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{pqr} + \frac{(r - 1)}{pqr} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{pqr} + \frac{1}{pq} - \frac{1}{pqr} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{pq} \\ \\ \\ hope \: it \: will \: help \: you..$