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Question: If T<sub>p</sub> = \(\frac { 1 } { q }\) are T<sub>q</sub> = \(\frac { 1 } { \mathrm { p } }\) o...

If Tp = 1q\frac { 1 } { q } are Tq = 1p\frac { 1 } { \mathrm { p } } of an A.P. then sum of (pq)th term =

A

pq−12\frac { p q - 1 } { 2 }

B
C

pq+12\frac { p q + 1 } { 2 }

D

−pq+12\frac { - \mathrm { pq } + 1 } { 2 }

Answer

pq+12\frac { p q + 1 } { 2 }

Explanation

Solution

A + (p – 1) d = 1/q

A + (q – 1) d = 1/p

(p – q) d = – 1p\frac { 1 } { \mathrm { p } }

̃ d =, A =

Spq = pq/2 [2A + (pq – 1) d]

= pq/2 [2pq+pq−1pq]\left[ \frac { 2 } { \mathrm { pq } } + \frac { \mathrm { pq } - 1 } { \mathrm { pq } } \right] =