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Question: If the \(p ^ { t h }\) term of an A.P. be \(\frac { 1 } { q }\) and <img src="https://cdn.pureessenc...

If the pthp ^ { t h } term of an A.P. be 1q\frac { 1 } { q } and term be1p\frac { 1 } { p }, then the sum of its terms will be.

A

pq−12\frac { p q - 1 } { 2 }

B

1−pq2\frac { 1 - p q } { 2 }

C

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

D

−pq+12- \frac { p q + 1 } { 2 }

Answer

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

Explanation

Solution

Since Tp=a+(p−1)d=1qT _ { p } = a + ( p - 1 ) d = \frac { 1 } { q } …..(i)

and Tq=a+(q−1)d=1pT _ { q } = a + ( q - 1 ) d = \frac { 1 } { p } …..(ii)

From (i) and (ii), we get a=1pqa = \frac { 1 } { p q } and d=1pqd = \frac { 1 } { p q }

Now sum of pqp q terms =pq2[2pq+(pq−1)1pq]= \frac { p q } { 2 } \left[ \frac { 2 } { p q } + ( p q - 1 ) \frac { 1 } { p q } \right]

=pq2⋅2pq[1+12(pq−1)]=[2+pq−12]=pq+12= \frac { p q } { 2 } \cdot \frac { 2 } { p q } \left[ 1 + \frac { 1 } { 2 } ( p q - 1 ) \right] = \left[ \frac { 2 + p q - 1 } { 2 } \right] = \frac { p q + 1 } { 2 }

Note : Students should remember this question as a formula.